RD SHARMA Solutions for Class 10 Maths Chapter 5 - Arithmetic Progressions

Chapter 5 - Arithmetic Progressions Exercise Ex. 5.1

Question 1 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (vi)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (vi)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (vii)

Write the first five terms for the sequence whose nth terms is:

an = n2 - n + 1

Solution 1 (vii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (viii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (viii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (ix)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (ix)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 2 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 2 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 2 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 2 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 2 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 2 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 2 (iv)

Find the indicated terms in the sequence whose nth terms are :

an = (n - 1) (2 - n) (3 + n); a1, a2, a3

Solution 2 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 2 (v)

Find the indicated terms in the sequence whose nth terms are:

an = (-1)n n; a3, a5, a8

Solution 2 (v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 3 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 3 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 3 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 3 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 3 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 3 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 3 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 3 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Chapter 5 - Arithmetic Progressions Exercise Ex. 5.2

Question 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 4 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 4 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 4 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 4 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 4 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 4 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 4 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 4 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 6(i)

Justify whether it is true to say that the sequence having following nth term is an A.P.

 

an = 2n - 1

Solution 6(i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 6(ii)

Justify whether it is true to say that the sequence having following nth term is an A.P.

 

an = 3n2 + 5

Solution 6(ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 6(iii)

Justify whether it is true to say that the sequence having following nth term is an A.P.

 

an = 1 + n + n2

Solution 6(iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Chapter 5 - Arithmetic Progressions Exercise Ex. 5.3

Question 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 2

Write the arithmetic progression when first term a and common difference d are as follows:

(i) a = 4, d = -3

(ii) a = -1, d = 1/2

(iii) a = -1.5, d = -0.5

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 3 (i)

In which of the following situations, the sequence of numbers formed will form an A.P.?

The cost of digging a well for the first metre is Rs 150 and rises by Rs 20 for each succeeding metre.


Solution 3 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 3 (ii)

In which of the following situations, the sequence of numbers formed will form an A.P.?

The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of their remaining in the cylinder.

Solution 3 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 3(iii)

In which of the following situations, the sequence of numbers formed will form an A.P.?

 

Divya deposited Rs.1000 at compound interest at the rate of 10% per annum. The amount at the end of first year, second year, third year, …., and so on.

Solution 3(iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 4 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 4 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 4 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 4 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 4 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 4 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 4 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 4 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5 (v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (vi)

Find out whether of the given sequence is an arithemtic progressions. If it is an arithmetic progressions, find out the common difference.

p, p + 90, p + 180, p + 270, ... where p = (999)999

Solution 5 (vi)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (vii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5 (vii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (viii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5 (viii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (ix)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5 (ix)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (x)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5 (x)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (xi)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5 (xi)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (xii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5 (xii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 6

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 6

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions
Question 7

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 7

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Chapter 5 - Arithmetic Progressions Exercise Ex. 5.4

Question 1 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (vi)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (vi)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (vii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (vii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 2 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 2 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 2 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 2 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 2 (iii)

Which term of the A.P. 4, 9, 14, ... is 254?

Solution 2 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 2 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 2 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 2 (v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 2 (v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 2 (vi)

Which term of the A.P. -7, -12, -17, -22,… will be -82? Is -100 any term of the A.P.?

Solution 2 (vi)

The given A.P. is -7, -12, -17, -22,…

First term (a) = -7

Common difference = -12 - (-7) = -5

Suppose nth term of the A.P. is -82.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

So, -82 is the 16th term of the A.P.

To check whether -100 is any term of the A.P., take an as -100.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

So, n is not a natural number.

Hence, -100 is not the term of this A.P.

Question 3 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 3 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 3 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 3 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 3 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 3 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 4 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 4 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 4 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 4 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 4 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 4 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 4 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 4 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 6

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 6

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 7

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 7

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 8

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 8

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 9 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 9 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 9 (ii)

Find the 12th term from the end of the following arithmetic progressions:

3, 8, 13, ..., 253

Solution 9 (ii)

A.P. is 3, 8, 13, ..., 253

We have:

Last term (l) = 253

Common difference (d) = 8 - 3 = 5

Therefore,

12th term from end

       = l - (n - 1)d

       = 253 - (12 - 1) (5)

       = 253 - 55

       = 198

Question 9 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 9 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 10

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 10

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 11

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 11

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 12

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 12

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 13

The 26th, 11th and last term of an A.P. are 0, 3 and Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions, respectively. Find the common difference and the number of terms.

Solution 13

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 14

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 14

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 15

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 15

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 16

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 16

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 17

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 17

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 18

The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of the A.P.

Solution 18

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 19

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 19

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 20 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 20 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 20 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 20 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 21

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 21

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions
Question 22 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 22 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 22 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 22 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 22 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 22 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 22 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 22 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 23

The eighth term of an A.P. is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.

Solution 23

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 24

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 24

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 25

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 25

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 26

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 26

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 27

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 27

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 28

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 28

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 29

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 29

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 30

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 30

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 31

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 31

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 32

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 32

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 33

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 33

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 34

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 34

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 35

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 35

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 36

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 36

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 37

The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, find the nth term.

Solution 37

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Thus, nth term is given by

an = a + (n - 1)d
an = 3 + (n - 1)4
an = 3 + 4n - 4
an = 4n - 1


Question 38

Find the number of all three digit natural numbers which are divisible by 9.

Solution 38

The smallest three digit number divisible by 9 = 108

The largest three digit number divisible by 9 = 999

 

Here let us write the series in this form,

108, 117, 126, …………….., 999

a = 108, d = 9

 

tn = a + (n - 1)d

999= 108 + (n - 1)9

999 - 108 = (n - 1)9

891 = (n - 1)9

(n - 1) = 99

n = 99 + 1

n = 100

Number of terms divisible by 9

 

Number of all three digit natural numbers divisible by 9 is 100.

Question 39

The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.

Solution 39

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

 

Question 40

The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P.

Solution 40

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

 

Question 41

The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is 4 times its 15th term.

Solution 41

Let the first term be 'a' and the common difference be 'd'

t24 = a + (24 - 1)d = a + 23d

t10­ = a + (10 - 1)d = a + 9d

t72 = a + (72 - 1)d = a + 71d

t15 = a + (15 - 1)d = a + 14d

 

t24 = 2t10

a + 23d = 2(a + 9d)

a + 23d = 2a + 18d

23d - 18d = 2a - a

5d = a

 

t72 = a + 71d

= 5d + 71d

= 76d

= 20d + 56d

= 4 × 5d + 4 × 14d

= 4(5d + 14d)

= 4(a + 14d)

= 4t15

t72 = 4t15

 

Question 42

Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.

Solution 42

L.C.M. of 2 and 5 = 10

3- digit number after 100 divisible by 10 = 110

3- digit number before 999 divisible by 10 = 990

 

Let the number of natural numbers be 'n'

990 = 110 + (n - 1)d

990 - 110 = (n - 1) × 10

880 = 10 × (n - 1)

n - 1 = 88

n = 89

 

The number of natural numbers between 110 and 999 which are divisible by 2 and 5 is 89.

Question 43

If the seventh term of an A.P. is 1/9 and its ninth term is 1/7, find its (63)rd term.

Solution 43

Let the first term be 'a' and the common difference be 'd'.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

 Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

 

 Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

 

Question 44

The sum of 5th and 9th terms of an A.P. is 30. If its 25th term is three times its 8th term, find the A.P.

Solution 44

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 45

Find where 0(zero) is a term of the A.P. 40, 37, 34, 31, ...

Solution 45

Let the first term be 'a' and the common difference be 'd'.

a = 40

d = 37 - 40 = - 3

Let the nth term of the series be 0.

tn = a + (n - 1)d

0 = 40 + (n - 1)( - 3)

0 = 40 - 3(n - 1)

3(n - 1) = 40

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

No term of the series is 0.

Question 46

Find the middle term of the A.P. 213, 205, 197, …, 37.

Solution 46

Given A.P. is 213, 205, 197, …, 37.

Here, first term = a = 213

And, common difference = d = 205 - 213 = -8

an = 37

nth term of an A.P. is given by

a­n = a + (n - 1)d

37 = 213 + (n - 1)(-8)

37 = 213 - 8n + 8

37 = 221 - 8n

8n = 221 - 37

8n = 184

n = 23

So, there are 23 terms in the given A.P.

The middle term is 12th term.

a12 = 213 + (12 - 1)(-8)

= 213 + (11)(-8)

= 213 - 88

= 125

Hence, the middle term is 125.

Question 47

If the 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term, find the A.P.

Solution 47

Let a be the first term and d be the common difference of the A.P.

Then, we have

a5 = 31 and a25 = a5 + 140

a + 4d = 31 and a + 24d = a + 4d + 140

a + 4d = 31 and 20d = 140

a + 4d = 31 and d = 7

a + 4(7) = 31 and d = 7

a + 28 = 31 and d = 7

a = 3 and d = 7

Hence, the A.P. is a, a + d, a + 2d, a + 3d, ……

i.e. 3, 3 + 7, 3 + 2(7), 3 + 3(7), ……

i.e. 3, 10, 17, 24, …..

Question 48

Find the sum of two middle terms of the A.P.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 48

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 49

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 49

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 50

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 50

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 51

How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?

Solution 51

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 52

Find the 12th term from the end of the A.P. -2, -4, -6, …, -100.

Solution 52

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 53

For the A.P.: -3, -7, -11, …, can we find a30 - a20 without actually finding a30 and a20? Give reason for your answer.

Solution 53

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 54

Two A.P.s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why?

Solution 54

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Chapter 5 - Arithmetic Progressions Exercise Ex. 5.5

Question 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 6

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 6

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 7

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 7

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 8

Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6.

Solution 8

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 9

The sum of the first three numbers in an arithmetic progression is 18. If the product of the first and third term is 5 times the common difference, find the three numbers.

Solution 9

Let the first three terms of an A.P. be a - d, a, a + d

As per the question,

(a - d) + a + (a + d) = 18

3a = 18

a = 6

Also, (a - d)(a + d) = 5d

(6 - d)(6 + d) = 5d

36 - d2 = 5d

d2 + 5d - 36 = 0

d2 + 9d - 4d - 36 = 0

(d + 9)(d - 4) = 0

d = -9 or d = 4

Thus, the terms will be 15, 6, -3 or 2, 6, 10.

Question 10

Spilt 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623.

Solution 10

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 11

The angles of a triangle are in A.P. the greatest angle is twice the least. Find all the angles.

Solution 11

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 12

The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the number.

Solution 12

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Chapter 5 - Arithmetic Progressions Exercise Ex. 5.6

Question 1 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (vi)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (vi)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (vii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (vii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 1 (viii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 1 (viii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 4

Find the sum of last ten terms of the A.P: 8, 10, 12, 14,.., 126.

Solution 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 5 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 5 (iv)

Find the sum of the first 15 terms of each of the following sequences having nth term as

yn = 9 - 5n

Solution 5 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 6

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 6

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 7

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 7

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 8

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 8

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 9

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 9

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 10 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 10 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 10 (ii)

How many terms are there in the A.P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40?

Solution 10 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 10 (iii)

How many terms of the A.P. 9, 17, 25, ... must be taken so that their sum is 636?

 

Remark* - Question modified.

Solution 10 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 10 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 10 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 10(v)

How many terms of the A.P. 27, 24, 21… should be taken so that their sum is zero?

Solution 10(v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 10 (vi)

How many terms of the A.P. 45, 39, 33 … must be taken so that their sum is 180? Explain the double answer.

Solution 10 (vi)

Let the required number of terms be n.

As the given A.P. is 45, 39, 33 …

Here, a = 45 and d = 39 - 45 = -6 

The sum is given as 180

Sn = 180

 Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

When n = 10,

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

When n = 6,

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Hence, number of terms can be 6 or 10.

Question 11 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 11 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 11 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 11 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 11 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 11 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 12(i)

Find the sum of

The first 15 multiples of 8.

Solution 12(i)

Multiples of 8 are8,16,24,…

Now,

n=15, a=8, d=8

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 12(ii)

Find the sum of

The first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

Solution 12(ii)

a) divisible by 3 3,6,9,… Now, n=40, a=3, d=3Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

 

b) divisible by 5 5,10,15,… Now, n=40, a=5, d=5Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

 

c)divisible by 6 6,12,18,… Now, n=40, a=3, d=6Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 12(iii)

Find the sum of

All 3 - digit natural numbers which are divisible by 13.

Solution 12(iii)

Three-digit numbers divisible by 13 are 104,117, 130,…988. Now, a=104, l=988Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 12(iv)

Find the sum of

All 3 - digit natural numbers, which are multiples of 11. 

Solution 12(iv)

Three-digit numbers which are multiples of 11 are 110,121, 132,…990. Now, a=110, l=990Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 12(v)

Find the sum of

All 2 - digit natural numbers divisible by 4. 

Solution 12(v)

Two-digit numbers divisible by 4 are 12,16,…96. Now, a=12, l=96Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 12 (vi)

Find the sum of first 8 multiples of 3.

Solution 12 (vi)

The multiples of 3 are 3, 6, 9, 12, 15, 18, 21, …

These are in A.P. with,

first term (a) = 3 and common difference (d) = 3

To find S8 when a = 3, d = 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, the sum of first 8 multiples of 3 is 108.

Question 13 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 13 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 13 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 13 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 13 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 13 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 13 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 13 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 13 (v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 13 (v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 13 (vi)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 13 (vi)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 13 (vii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 13 (vii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 13 (viii)

Find the sum:

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 13 (viii)

Let 'a' be the first term and 'd' be the common difference.

 Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

tn = a + (n - 1)d

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

 

Question 14

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 14

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 15

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 15

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 16

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 16

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 17

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 17

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 18

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 18

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 19

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 19

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 20

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 20

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 21

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 21

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 22 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 22 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 22 (ii)

If the sum of first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the sum of its first n terms.

Solution 22 (ii)

Sum of first n terms of an AP is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

As per the question, S4 = 40 and S14 = 280

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Also, Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Subtracting (i) from (ii), we get, 10d = 20

Therefore, d = 2

Substituting d in (i), we get, a = 7

Sum of first n terms becomes

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Question 23

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 23

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 24

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 24

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 25

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 25

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 26

The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.

Solution 26

Let the number of terms be 'n', 'a' be the first term and 'd' be the common difference.

tn = a + (n - 1)d

49 = 7 + (n - 1)d

42 = (n - 1)d…..(i)

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

 

840 = n[14 + (n - 1)d]……(ii)

 

Substituting (ii) in (i),

840 = n[14 + 42]

840 = 56n

n = 15

 

Substituting n in (i)

42 = (15 - 1)d

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Common difference, d =3

Question 27

The first and the last terms of an A.P. are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.

Solution 27

Let the number of terms be 'n', 'a' be the first term and 'd' be the common difference.

tn = a + (n - 1)d

45 = 5 + (n - 1)d

40 = (n - 1)d…..(i)

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

 

800 = n[10 + (n - 1)d]……(ii)

 

Substituting (ii) in (i),

800 = n[10 + 40]

800 = 50n

n = 16

 

Substituting n in (i)

40 = (16 - 1)d

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 28

The sum of first q terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1 : 2. Find the first and 15th term of the A.P.

Solution 28

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Question 29

If the 10th term of an A.P. is 21 and the sum of its first ten terms is 120, find its nth term.

Solution 29

Let 'a' be the first term and 'd' be the common difference.

tn = a + (n - 1)d

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

 

t10 = a + (10 - 1)d

21 = a + 9d……(i)

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

120 = 5[2a + 9d]

24 = 2a + 9d………(ii)

 

(ii) - (i)  

a = 3

Substituting a in (i), we get

a + 9d = 21

3 + 9d = 21

9d = 18

d = 2

 

tn = a + (n - 1)d

= 3 + (n - 1)2

= 3 + 2n - 2

tn= 2n + 1

 

Question 30

The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P.

Solution 30

Let the first term be 'a' and the common difference be 'd' Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

 

63 = 7[a + 3d]

9 = a + 3d……….(i)

 

Sum of the next 7 terms = 161

Sum of the first 14 terms = 63 + 161 = 224

 Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

224 = 7[2a + 13d]

32 = 2a + 13d………..(ii)

 

Solving (i) and (ii), we get

d = 2, a = 3

 

28th term of the A.P., t28 = a + (28 - 1)d

= 3 + 27 × 2

= 3 + 54

= 57

 

The 28th of the A.P. is 57.

Question 31

The sum of first seven terms of an A.P. is 182. If its 4th and the 17th terms are in the ratio 1: 5, find the A.P.

Solution 31

Let the first term be 'a' and the common difference be 'd'.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

 

26 × 2 = [2a + (7 - 1)d]

52 = 2a + 6d

26 = a + 3d……..(i)

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

From (i) and (ii),

13d = 104

d = 8

 

From (i), a = 2

 

The A.P. is 2, 10, 18, 26,…….

Question 32

The nth term of an A.P. is given by (- 4n + 15). Find the sum of the first 20 terms of this A.P.

Solution 32

Let the first term of the A.P. be 'a' and the common difference be 'd'.

 

tn = - 4n + 15

t1 = - 4 × 1 + 15 = 11

t2 = - 4 × 2 + 15 = 7

t3 = - 4 × 3 + 15 = 3

 

Common Difference, d = 7 - 11 = -4

 

 Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

= 10 × (-54)

= - 540

 

*Note: Answer given in the book is incorrect.

Question 33

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 33

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 34

Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term.

Solution 34

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 35

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 35

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 36

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 36

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 37

Find the number of terms of the A.P. - 12, - 9, - 6, …, 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.

Solution 37

Let the number of the terms be 'n'.

Common Difference, d = - 9 + 12 = 3

 

tn = a + (n - 1)d

21 = - 12 + 3(n - 1)

21 + 12 = 3(n - 1)

3(n - 1) = 33

n - 1 = 11

n = 12

 

Number of terms of the series = 12

 

If 1 is added to each term of the above A.P.,

- 11, - 8, - 5,…….,22

Number of terms in the series, n1 = 12

Sum of all the terms,

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

The sum of the terms = 66

Question 38

The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P.

Solution 38

Sum of n terms of the A.P., Sn = 3n2 + 6n

S1 = 3 × 12 + 6 × 1 = 9 = t1 ……(i)

S2 = 3 × 22 + 6 × 2 = 24 = t1 + t2 …….(ii)

S3 = 3 × 32 + 6 × 3 = 45 = t1 + t2 + t3 ……..(iii)

 

From (i), (ii) and (iii),

t1 = 9, t2 = 15, t3 = 21

 

Common difference, d = 15 - 9 = 6

nth of the AP, tn = a + (n - 1)d

= 9 + (n - 1) 6

= 9 + 6n - 6

= 6n + 3

 

Thus, the nth term of the given A.P. = 6n + 3

Question 39

The sum of the first n terms of an A.P. is 5n - n2. Find the nth term of this A.P.

Solution 39

Sn = 5n - n2

S1 = 5 × 1 - 12 = 4 = t1………..(i)

S2 = 5 × 2 - 22 = 6 = t1 +t2………..(ii)

S3 = 5 × 3 - 32 = 6 = t1 + t2 + t……….(iii)

 

From (i), (ii) and (iii),

t1 = 4, t2 = 2, t3 = 0

 

Here a = 4, d = 2 - 4 = - 2

tn = a + (n - 1)d

= 4 + (n - 1)( -2)

= 4 - 2n + 2

= 6 - 2n

Question 40

The sum of the first n terms of an A.P. is 4n2 + 2n. find the nth term of this A.P.

Solution 40

Sn = 4n2 + 2n

S1 = 4 × 12 + 2 × 1 = 6 = t1………….(i)

S2 = 4 × 22 + 2 × 2 = 20 = t1 + t2……….(ii)

S3 = 4 × 32 + 2 × 3 = 42 = t1 + t2 + t3………..(iii)

 

From (i), (ii) and (iii),

t1 = 6, t2 = 14, t3 = 22

 

Here a = 6, d = 14 - 6 = 8

tn = a + (n - 1)d

tn = 6 + (n - 1)8

= 6 + 8n - 8

= 8n - 2

 

*Note: Answer given in the book is incorrect.

 

Question 41

The sum of first n terms of an A.P. is 3n2 + 4n. find the 25th term of this A.P.

Solution 41

Sum of n terms of the A.P., Sn = 3n2 + 4n

S1 = 3 × 12 + 4 × 1 = 7 = t1………(i)

S2 = 3 × 22 + 4 × 2 = 20 = t1 + t…….(ii)

S3 = 3 × 32 + 4 × 3 = 39 = t1 + t2 + t3 …….(iii)

 

From (i), (ii), (iii)

t1 = 7, t2 = 13, t­3 = 19

Common difference, d = 13 - 7 = 6

 

25th of the term of this A.P., t25 = 7 + (25 - 1)6

= 7 + 144 = 151

The 25th term of the A.P. is 151.

Question 42

The sum of first n terms of an A.P. is 5n2 + 3n. If its mth term is 168, find the value of m. Also, find the 20th term of this A.P.

Solution 42

Sum of the terms, Sn = 5n2 + 3n

S1 = 5 × 12 + 3 × 1 = 8 = t1………..(i)

S2 = 5 × 22 + 3 × 2 = 26 = t1 + t2…………..(ii)

S3 = 5 × 32 + 3 × 3 = 54 = t1 + t2 + t3…………(iii)

 

From(i), (ii) and (iii),

t1 = 8, t2 = 18, t3 = 28

Common difference, d = 18 - 8 = 10

 

tm = 168

a + (m - 1)d = 168

8 + (m - 1)×10 = 168

(m - 1) × 10 = 160

m - 1 = 16

m = 17

t20 = a + (20 - 1)d

= 8 + 19 × 10

= 8 + 190

= 198

 

Question 43

The sum of first q terms of an A.P. is 63q - 3q2. If its pth term -60, find the value of p. Also, find the 11th term of this A.P.



Remark* - Question modified.

Solution 43

Let the first term be 'a' and the common difference be 'd'.

Sum of the first 'q' terms, Sq = 63q - 3q2

S1 = 63 × 1 - 3 × 12 = 60 = t1……..(i)

S2 = 63 × 2 - 3 × 22 = 114 = t1 + t2…..(ii)

S3 = 63 × 3 - 3 × 32 = 162 = t1 + t2 + t3 .....(iii)

From (i), (ii) and (iii),

t1 = 60

t2 = 54

t3 = 48

 

Common difference, d = 54 - 60 = - 6

 

tp = a + (p - 1)d

-60 = 60 + (p - 1)( - 6)

- 120 = - 6(p - 1)

p - 1 = 20

p = 21

 

t11 = 60 + (11 - 1)( - 6)

=60 + 10( - 6)

= 60 - 60

= 0

The 11th term of the A.P. is 0.

Question 44

The sum of first m terms of an A.P. is 4m2 - m. If its nth term is 107, find the value of n. Also, find the 21st term of this A.P.

Solution 44

Let the first term of the A.P. be 'a' and the common difference be 'd'

 

Sum of m terms of the A.P., Sm = 4m2 - m

S1 = 4 × 12 - 1 = 3 = t1 …….(i)

S2 = 4 × 22 - 2 = 14 = t1 + t2……..(ii)

S3 = 4 × 32 - 3 = 33 = t1 + t2 + t …….(iii)

 

From (i), (ii) and (iii)

t1 = 3, t2 = 11, t3 = 19

Common difference, d = 11 - 3 = 8

 

tn = 107

a + (n - 1)d = 107

3 + (n - 1)8 = 107

8(n - 1) = 104

n - 1 = 13

n = 14

 

t21 = 3 + (21 - 1)8 = 3 + 160 = 163

Question 45

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 45

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 46

If the sum of first n terms of an A.P. is Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressionsthen find its nth term. Hence write its 20th term.

Solution 46

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 47 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 47 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 47 (ii)

If the sum of first n terms of an A.P. is n2, then find its 10th term.

Solution 47 (ii)

Sum of first n terms of an AP is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

As per the question, Sn = n2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, the 10th term of this A.P. is 19.

Question 48

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 48

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 49

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 49

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 50 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 50 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 50 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 50 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 51

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 51

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 52

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 52

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 53

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 53

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 54

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 54

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 55 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 55 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 55(ii)

Find the sum of all integers between 100 and 550 which are not divisible by 9.

Solution 55(ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 55(iii)

Find the sum of all integers between 1 and 500 which are multiplies 2 as well as of 5.

Solution 55(iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 55(iv)

Find the sum of all integers from 1 to 500 which are multiplies 2 as well as of 5.

Solution 55(iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 55(v)

Find the sum of all integers from 1 to 500 which are multiples of 2 or 5.

Solution 55(v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 56 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 56 (i)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 56 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 56 (ii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 56 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 56 (iii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 56 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 56 (iv)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 56 (v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 56 (v)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 56 (vi)

Let there be an A.P. with first term 'a', common difference 'd'. If an denotes its nth term and Sn the sum of first n terms, find

n and an, if a = 2, d = 8 and Sn = 90.

Solution 56 (vi)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 56(vii)

Let there be an A.P. with first term 'a', common difference 'd'. If an denotes its nth term and Sn the sum of first n terms, find k, if Sn = 3n2 + 5n and ak = 164.

Solution 56(vii)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 56 (viii)

Let there be an A.P. with first term 'a', common difference 'd'. If an denotes its nth term and Sn the sum of first n terms, find S22, if d = 22 and a22 = 149.

Solution 56 (viii)

The nth term of an A.P. is given by an = a + (n - 1)d

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Sum of first n terms of an AP is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 57

If Sn denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 - S4).

Solution 57

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 58

A thief, after committing a theft runs at a uniform speed of 50 m/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5 m/minute every succeeding minute. After how many minutes, the policeman will catch the thief?

Solution 58

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 59

The sums of first n terms of three A.P.s are S1, S2 and S3. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that S1 + S3 = 2S2.

Solution 59

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 60

Resham wanted to save at least Rs. 6500 for sending her daughter to school next year (after 12 months). She saved Rs. 450 in the first month and raised her saying by Rs.20 every next month. How much will she be able to save in next 12 months? Will she be able to send her daughter to the school next year?

Solution 60

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 61

In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students.

Solution 61

Trees planted by the student in class 1 = 2 + 2 = 4

Trees planted by the student in class 2 = 4 + 4 = 8

Trees planted by the students in class 3 = 6 + 6 = 12

…….

Trees planted by the students in class 12 = 24 + 24 = 48

 

the series will be 4, 8, 12,………., 48

a = 4, Common Difference, d = 8 - 4 = 4

Let 'n' be the number of terms in the series.

48 = 4 + (n - 1)4

44 = 4(n - 1)

n - 1 = 11

n = 12

Sum of the A.P. series,

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Number of trees planted by the students = 312

Question 62

Ramkali would need Rs. 1800 for admission fee and books etc., for her daughter to start going to school from the next year. She saved Rs. 50 in the first month of this year and increased her monthly saving by Rs. 20. After a year, how much money will she save? Will she be able to fulfill her dream of sending her daughter to school?

Solution 62

Since, the difference between the savings of two consecutive months is Rs. 20, therefore the series is an A.P.

 

Here, the savings of the first month is Rs. 50

First term, a = 50, Common difference, d = 20

No. of terms = no. of months

No. of terms, n = 12

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

= 6[100 + 220]

= 6×320

= 1920

 

After a year, Ramakali will save Rs. 1920.

Yes, Ramakali will be able to fulfill her dream of sending her daughter to school.

Question 63

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 63

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 64

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 64

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 65

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 65

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 66

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 66

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 67

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 67

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 68

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 68

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 69

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Solution 69

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 70

If Sn denotes the sum of the first n terms of an A.P., prove that S30 = 3(S20 - S10).

Solution 70

Let the first term of the A.P. be 'a' and the common difference be 'd'.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

 

R.H.S.

= 3(S20 - S10)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

= 3(10[2a + 19d] - 5[2a + 9d])

= 3(20a + 190d - 10a - 45d)

= 3(10a + 145d)

= 3 × 5(2a + 29d)

= 15[2a + (30 - 1)d]

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

= S30

= L.H.S.

Question 71

Solve the equation

(-4) + (-1) + 2 + 5 + …. + x = 437.

Solution 71

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 72

Which term of the A.P. -2, -7, -12,…, will be -77? Find the sum of this A.P. upto the term -77.

Solution 72

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 73

The sum of the first n terms of an A.P. whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another A.P. whose first term is -30 and common difference is 8. Find n.

Solution 73

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Question 74

The students of a school decided to beautify the school on the annual day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 metre. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?

Solution 74

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Chapter 5 - Arithmetic Progressions Exercise Ex. MCQs

Question 1

If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is

(a) 87 

(b) 88

(c) 89

(d) 90

Solution 1

The nth term of an A.P. is given by an = a + (n - 1)d

As per the question, a7 = 34 and a13 = 64

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Also, Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Subtracting (i) from (ii), we get, 6d = 30

Therefore, d = 5

Substituting a in (i), we get, a = 4

The 18th term is

a18 = a + 17d

 = 4 + 85

 = 89

Thus, the 18th term is 89.

Hence, option (c) is correct.

Question 2

If the sum of p terms of an A.P. is q and the sum of q terms is p, then the sum of p + q terms will be

(a) 0 

(b) p - q

(c) p + q

(d) -(p + q)

Solution 2

The sum of first n terms of an A.P. is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

As per the question, Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Subtracting (ii) from (i), we get

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Now, sum of (p + q) terms is

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (d) is correct.

Question 3

If the sum of n terms of an A.P. be 3n2 + n and its common difference is 6, then its first term is

(a) 2 

(b) 3

(c) 1

(d) 4

Solution 3

Given: Sn = 3n2 + n and d = 6

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Thus, the first term is 4.

Hence, option (d) is correct.

Question 4

The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be

(a) 5

(b) 6

(c) 7

(d) 8

Solution 4

Given: First term (a) = 1, last term (l) = 11 and Sn = 36

We know that, sum of n terms is given by Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Thus, the number of terms is 6.

Hence, option (b) is correct.

Question 5

If the sum of n terms of an A.P. is 3n2 + 5n then which of its terms is 164?

(a) 5

(b) 6

(c) 7

(d) 8

Solution 5

Given: Sum of n terms Sn = 3n2 + 5n

For n = 1, we get

S1 = 3(1)2 + 5(1) = 8

For n = 2, we get

S2 = 3(2)2 + 5(2) = 22

We know that, an = Sn - Sn - 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Therefore, we have

Common difference (d) = 14 - 8 = 6

Let 164 be the nth term of this A.P.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Thus, 164 is the 27th term of this A.P.

Hence, option (b) is correct.

Question 6

If the sum of n terms of an A.P. is 2n2 + 5n, then its nth term is

(a) 4n - 3

(b) 3n - 4

(c) 4n + 3

(d) 3n + 4

Solution 6

Given: Sum of n terms Sn = 2n2 + 5n

For n = 1, we get

S1 = 2(1)2 + 5(1) = 7

For n = 2, we get

S2 = 2(2)2 + 5(2) = 18

We know that, an = Sn - Sn - 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Here, common difference (d) = 11 - 7 = 4

The nth term is given by

an = a + (n - 1)d

 = 7 + (n - 1)(4)

 = 4n + 3

Thus, the nth term is 4n + 3.

Hence, option (c) is correct.

Question 7

If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is

(a) 13

(b) 9

(c) 21

(d) 17

Solution 7

Let (a - d), a and (a + d) be the first three consecutive terms of an A.P.

As per the question, we have

(a - d) + a + (a + d) = 51

i.e. 3a = 51

i.e. a = 17

Also, (a - d)(a + d) = 273

(a2 - d2) = 273

289 - d2 = 273

d2 = 289 - 273

d2 = 16

d = ± 4

As the series is an increasing A.P., d must be positive.

Therefore, d = 4

So, we get, a + d = 21

Hence, option (c) is correct.

Question 8

If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are

(a) 5, 10, 15, 20

(b) 4, 10, 16, 22

(c) 3, 7, 11, 15

(d) none of these

Solution 8

Let (a - 3d), (a - d), (a + d) and (a + 3d) be the four numbers of an A.P.

As per the question, we have

(a - 3d) + (a - d) + (a + d) + (a + 3d) = 50

i.e. 4a = 50

i.e. a = 12.5

Also, a + 3d = 4(a - 3d)

i.e. a + 3d = 4a - 12d

i.e. 3a = 15d

i.e. a = 5d

i.e. 5d = 12.5

Therefore, d = 2.5

So, (a - 3d) = 5, (a - d) = 10, (a + d) = 15 and (a + 3d) = 20.

Thus, the numbers are 5, 10, 15 and 20.

Hence, option (a) is correct.

Question 9

Let Sn denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = Sn - kSn-1 + Sn-2,  then k =

(a) 1

(b) 2

(c) 3

(d) none of these

Solution 9

Let a be the first term.

Sum of n terms of an A.P. is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

So, the sum of (n - 1) terms is

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

And, sum of (n - 2) terms is

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

As it is given that d = Sn - kSn-1 + Sn-2, we have

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Hence, option (b) is correct.

Question 10

The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by  Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions then k =  

(a) S

(b) 2S

(c) 3S

(d) none of these

Solution 10

Sum of n terms of an A.P. is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (b) is correct.

Question 11

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then k =  

(a) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(b) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(c) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(d) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Solution 11

The n even natural numbers 2, 4, 6,… forms an A.P. with first terms 2 and common difference 2.

So, the sum of first n even natural numbers is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

The n odd natural numbers 1, 3, 5,… forms an A.P. with first terms 1 and common difference 2.

So, the sum of first n odd natural numbers is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

As per the question,

Se = k So

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (b) is correct.

Question 12

If the first, second and last terms of an A.P. are a, b and 2a respectively, its sum is   

(a) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(b) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(c) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(d) none of these 

Solution 12

Given: First term = a, second term = b and last term = 2a

Therefore, common difference (d) = b - a

As the last term is 2a, we have

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

The of all the terms is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Hence, option (c) is correct.

Question 13

If S1 is the sum of an arithmetic progression of 'n' odd number of terms and S2 the sum of the terms of the series in odd places, then Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(a) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(b) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(c) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(d) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Solution 13

Sum of 'n' odd number of terms of an A.P. is

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Therefore, we have

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

From equations (i) and (ii), we get

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (a) is correct.

Question 14

If in an A.P., Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to

(a) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(b) mnp 

(c) p3 

(d) (m + n)p2 

Solution 14

We know that, sum of n terms of an A.P. is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Also, Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

From (i) and (ii), we get

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

From (i), we get

2a = 2np - (n - 1)2p

i.e. 2a = 2np - 2np +2p

i.e. a = p

Now, the sum of p terms Sp is

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (c) is correct.

Question 15

If Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn, then S3n : Sn is equal to

(a) 4 

(b) 6

(c) 8

(d) 10 

Solution 15

We know that, sum of n terms of an A.P. is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

As S2n = 3Sn, we have

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Consider,

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Thus, S3n : Sn = 6.

Hence, option (b) is correct.

Question 16

In an AP, Sp = q, Sq = p and Sr denotes the sum of first r terms. Then, Sp+q is equal to

(a) 0 

(b) -(p + q)

(c) p + q

(d) pq

Solution 16

Given: Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

The sum of first n terms of an A.P. is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Subtracting (ii) from (i), we get

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Therefore, Sp+q is

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (b) is correct.

Question 17

If Sr denotes the sum of the first r terms of an A.P. Then, S3n : (S2n - Sn) is

(a) n

(b) 3n

(c) 3

(d) none of these

Solution 17

Given: Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

The sum of first n terms of an A.P. is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Therefore, S3n will be

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Similarly,

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

From (i), (ii) and (iii), we have

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Thus, Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Hence, option (c) is correct.

Question 18

If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 terms is

(a) 3200

(b) 1600

(c) 200

(d) 2800

Solution 18

Given: First term (a) = 2 and common difference (d) = 4

The sum of first n terms of an A.P. is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Sum of its 40 terms will be

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (a) is correct.

Question 19

The number of terms of the A.P. 3, 7, 11, 15, … to be taken so that the sum is 406 is

(a) 5

(b) 10

(c) 12

(d) 14

Solution 19

In the given A.P. 3, 7, 11, 15, … we have

First term (a) = 3 and common difference (d) = 4

The sum of first n terms of an A.P. is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

As the sum is given as 406, we have

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Since the number of terms can't be negative, so n = 14.

Hence, option (d) is correct.

Question 20

Sum of n terms of the series Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions is 

(a) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(b) 2n(n + 1)

(c) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(d) 1

Solution 20

For the given series, the termsRd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions forms an A.P.

Here, first term (a) = Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions and common difference (d) = Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

The sum of first n terms of an A.P. is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Therefore, sum of the series is

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (c) is correct.

Question 21

The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to zero is

(a) 501th 

(b) 502th

(c) 458th 

(d) None of these

Solution 21

In an A.P., a9 = 449 and a449 = 9

We know that the nth term is given by

an = a + (n - 1)d

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Subtracting (ii) from (i), we get

-440d = 440

Therefore, d = -1

Putting the value of d in (i), we get

a - 8 = 449

Therefore, a = 457

Let nth term be 0 i.e. an = 0

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (c) is correct.

Question 22

If Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions are in A.P. Then, x =  

(a) 5 

(b) 3

(c) 1 

(d) 2

Solution 22

As Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions are in A.P.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (c) is correct.

Question 23

The nth term of an A.P., the sum of whose n terms is Sn, is

(a) Sn + Sn-1 

(b) Sn - Sn-1

(c) Sn + Sn+1 

(d) Sn - Sn+1

Solution 23

Let the A.P. be a1, a2, a3, …

As Sn is the sum of n terms of an A.P., we have

Sn = a1 + a2 + a3 + … + an

Sn-1 = a1 + a2 + a3 + … + an-1

Sn - Sn-1 = a1 + a2 + a3 + … + an-1 + an - (a1 + a2 + a3 + … + an-1)

 = an

Thus, Sn - Sn-1 = an.

Hence, option (b) is correct.

Question 24

The common difference of an A.P., the sum of whose n terms is Sn, is

(a) Sn - 2Sn-1 + Sn-2 

(b) Sn - 2Sn-1 - Sn-2

(c) Sn - Sn-2 

(d) Sn - Sn-1

Solution 24

Let the A.P. be a1, a2, a3, …

When Sn is the sum of n terms of an A.P., nth term will be

an = Sn - Sn-1

Similarly,

an-1 = Sn-1 - S(n-1) - 1 = Sn-1 - Sn-2

Now, the common difference is given by

d = an - an-1

 = Sn - Sn-1 - (Sn-1 - Sn-2)

 = Sn - 2Sn-1 + Sn-2

Hence, option (a) is correct.

Question 25

If the sums of n terms of two arithmetic progressions are in the ratio Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions then their nth terms are in the ratio 

(a) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(b) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(c) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(d) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Solution 25

Let Sn and S'n denotes the sum of first n terms of the two APs respectively.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Ratio of the sum of two APs is

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Replacing n with (2n - 1), we get

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (b) is correct.

Question 26

If Sn denote the sum of n terms of an A.P. with first term a and common d such that Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions is independent of x, then

(a) d = a 

(b) d = 2a 

(c) a = 2d 

(d) d = -a 

Solution 26

As Sn is the sum of n terms of an A.P., we have

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

As Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions is independent of x, then

2a - d = 0

i.e. 2a = d

Putting this in the ratio, we get

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Thus, 2a = d.

Hence, option (b) is correct.

Question 27

If the first term of an A.P. is a and nth term is b, then its common difference is 

(a) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(b) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(c) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(d) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Solution 27

The nth term of an A.P. is given by an = a + (n - 1)d

Here, 'a' and 'd' are first term and common difference

As b is the nth term of an A.P., we have

b = a + (n - 1)d

b - a = (n - 1)d

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (b) is correct.

Question 28

The sum of first n odd natural numbers is

(a) 2n - 1 

(b) 2n + 1 

(c) n2 

(d) n2 - 1 

Solution 28

The n odd natural numbers are 1, 3, 5, … n.

These terms forms an A.P. with first term (a) = 1 and common difference (d) = 2.

So, the sum of first n odd natural numbers will be

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (c) is correct.

Question 29

Two A.P.'s have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their 30th terms is

(a) 11 

(b) 3

(c) 8 

(d) 5 

Solution 29

Let d be the common difference of the two A.P.'s.

First term of 1st A.P. (a) = 8

First term of 2nd A.P. (a') = 3

Now, the 30th term of 1st A.P. is

a30 = a + (30 - 1)d

i.e. a30 = 8 + 29d … (i)

The 30th term of 2nd A.P. is

a'30 = a' + (30 - 1)d

a'30 = 3 + 29d … (ii)

Difference between these 30th terms is

a30 - a'30 = (8 + 29d) - (3 + 29d) = 5

Hence, option (d) is correct.

Question 30

If 18, a, b, -3 are in A.P., then a + b =

(a) 19

(b) 7

(c) 11 

(d) 15 

Solution 30

As 18, a, b, -3 are in A.P.

So, the difference between every two consecutive terms will be equal.

i.e. a - 18 = b - a = -3 - b

i.e. a - 18 = -3 - b

i.e. a + b = 18 - 3

i.e. a + b = 15

Hence, option (d) is correct.

Question 31

The sum of n terms of two A.P.'s are in the ratio 5n+9:9n+6. Then, the ratio of their 18th term is

(a) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(b) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(c) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(d) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Solution 31

Let Sn and S'n denotes the sum of first n terms of the two APs respectively.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Ratio of the sum of two APs is

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Replacing n with (2n - 1), we get

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

So, the ratio of their 18th terms is

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (a) is correct.

Question 32

If Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions then n =

(a) 8 

(b) 7 

(c) 10 

(d) 11 

Solution 32

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Here, 5, 9, 13, … forms an A.P. with first term (a) = 5 and common difference (d) = 4

Also, 7, 9, 11, … forms an A.P. with first term (a') = 7 and common difference (d') = 2

We know that, sum of n terms of an A.P. is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

And, Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

From (i), we get

 Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Since the number of terms can't be negative, so n = 7.

Hence, option (b) is correct.

Question 33

The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its

(a) 24th term

(b) 27th term 

(c) 26th term 

(d) 25th term 

Solution 33

Let the A.P. be a1, a2, a3, …

When Sn is the sum of n terms of an A.P., nth term will be

an = Sn - Sn-1

But, Sn = 3n2 + 5n

So, the nth term will be

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Let 164 be the nth term

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (b) is correct.

Question 34

If the nth term of an A.P. is 2n + 1, then the sum of first n terms of the A.P. is

(a) n(n - 2)

(b) n(n + 2)

(c) n(n + 1) 

(d) n(n - 1) 

Solution 34

The nth term of an A.P. is given as

an = 2n + 1

First term = a1 = 2(1) + 1 = 3

a2 = 2(2) + 1 = 5

So, the common difference (d) = a2 - a1 = 2

Now, the sum of n terms of an A.P. is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (b) is correct.

Question 35

If 18th and 11th term of an A.P. are in the ratio 3 : 2, then its 21st and 5th terms are in the ratio

(a) 3 : 2

(b) 3 : 1

(c) 1 : 3

(d) 2 : 3

Solution 35

The nth term of an A.P. is given by

an = a + (n - 1)d

Ratio of 18th and 11th terms is

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Now, the ratio of 21st and 5th terms is

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (b) is correct.

Question 36

The sum of first 20 odd natural numbers is

(a) 100

(b) 210

(c) 400

(d) 420

Solution 36

The n odd natural numbers are 1, 3, 5, … n.

These terms forms an A.P. with first term (a) = 1 and common difference (d) = 2.

So, the sum of first n odd natural numbers will be

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Now, the sum of first 20 odd natural numbers is

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (c) is correct.

Question 37

The common difference of the A.P. Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions is

(a) -1

(b) 1

(c) q

(d) 2q

Solution 37

The common difference of an A.P. with terms a1, a2, a3, … is given by

d = a2 - a1 = a3 - a2 = a4 - a3 = …

In the A.P. Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions we have

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Therefore, Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Hence, option (a) is correct.

Question 38

The common difference of the A.P. Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions is

(a) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(b) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(c) -b

(d) b

Solution 38

The common difference of an A.P. with terms a1, a2, a3, … is given by

d = a2 - a1 = a3 - a2 = a4 - a3 = …

In the given A.P. Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions we have

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Therefore, Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Hence, option (c) is correct.

Question 39

The common difference of the A.P. Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions is

(a) 2b 

(b) -2b 

(c) 3

(d) -3

Solution 39

The common difference of an A.P. with terms a1, a2, a3, … is given by

d = a2 - a1 = a3 - a2 = a4 - a3 = …

In the given A.P. Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions we have

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Therefore, Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Hence, option (d) is correct.

Question 40

If k, 2k - 1 and 2k + 1 are three consecutive terms of an AP, the value of k is

(a) -2

(b) 3

(c) -3

(d) 6

Solution 40

Given: k, 2k - 1 and 2k + 1 are three consecutive terms of an AP

Therefore, common difference (d) = (2k - 1) - k

Also, d = (2k + 1) - (2k - 1)

(2k - 1) - k = (2k + 1) - (2k - 1) 

k - 1 = 1 + 1

k = 3

Hence, option (b) is correct.

Question 41

The next term of the A.P. Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

(a) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(b) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(c) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

(d) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Solution 41

The given A.P. is Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

i.e. Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions 

Here, first term (a) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

The common difference (d) = Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

As the A.P. is given by a, (a + d), (a + 2d), (a + 3d), …

So, the next term will be

(a + 3d) = Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (d) is correct.

Question 42

The first three terms of an A.P. respectively are 3y - 1, 3y + 5 and 5y + 1. Then, y equals

(a) -3 

(b) 4 

(c) 5 

(d) 2 

Solution 42

As the terms 3y - 1, 3y + 5 and 5y + 1 are in A.P.

Therefore, we have

3y + 5 - (3y - 1) = 5y + 1 - (3y + 5)

3y + 5 - 3y + 1 = 5y + 1 - 3y - 5

6 = 2y - 4

y = 5

Hence, option (c) is correct.

Question 43

The sum of first five multiples of 3 is

(a) 45 

(b) 55 

(c) 65 

(d) 75 

Solution 43

The first five multiples of 3 are 3, 6, 9, 12, 15.

These terms form an A.P. with,

First term (a) = 3 and common difference (d) = 3

We know that, sum of first n terms of an A.P. is Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

As sum of first 5 multiples is same as sum of first 5 terms of an A.P., we have

Sum of first five multiples = S5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (a) is correct.

Question 44

The sum of first 16 terms of the A.P.: 10, 6, 2 …, is

(a) -320

(b) 320

(c) -352 

(d) -400

Solution 44

For the given A.P. 10, 6, 2, …, we have

First term (a) = 10 and common difference (d) = -4

We know that, sum of first n terms of an A.P. is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

So, the sum of first 16 terms will be

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (a) is correct.

Question 45

If the first term of an A.P. is -5 and the common difference is 2, then the sum of first 6 terms is

(a) 0

(b) 5

(c) 6

(d) 15

Solution 45

In the A.P., we have

First term (a) = -5 and common difference (d) = 2

We know that, sum of first n terms of an A.P. is given by

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

So, the sum of first 6 terms will be

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Arithmetic Progressions

Hence, option (a) is correct.

Question 46

The 4th term from the end of the AP: -11, -8, -5, …, 49 is

(a) 37

(b) 40

(c) 43

(d) 58

Solution 46

In the A.P., first term (a) = -11, last term (l) = 49 and common difference (d) = 3

Here, nth term from the end = l - (n - 1)d

Therefore,

4th term from the end = 49 - (4 - 1)(3)

= 49 - 9

= 40

Hence, option (b) is correct.

Question 47

Which term of the A.P. 21, 42, 63, 84, … is 210?

(a) 9th

(b) 10th

(c) 11th

(d) 12th

Solution 47

In the given A.P., first term (a) = 21 and common difference (d) = 21

Let 210 be the nth term of the AP.

So, we have

210 = a + (n - 1)d

210 = 21 + (n - 1)(21)

21(n - 1)= 189

n - 1 = 9

Therefore, n = 10

Hence, option (b) is correct.

Question 48

If the 2nd term of an A.P. is 13 and 5th term is 25, what is its 7th term?

(a) 30

(b) 33

(c) 37

(d) 38

Solution 48

Given: 2nd term = 13 and 5th term = 25

a + (2 - 1)d = 13 and a + (5 - 1)d = 25

a + d = 13 … (i) and,

a + 4d = 25 … (ii)

Subtracting (i) from (ii), we have

3d = 12

Therefore, d = 4

Substituting the value of 'd' in (i), we get

a = 9

So, the 7th term will be

a7 = a + 6d = 9 + 24 = 33

Hence, option (b) is correct.

Question 49

The value of x for which 2x, x + 10 and 3x + 2 are the three consecutive terms of an A.P; is

(a) 6

(b) -6

(c) 18

(d) -18

Solution 49

As 2x, x + 10 and 3x + 2 are the three consecutive terms of an A.P.

We have

x + 10 - 2x = 3x + 2 - (x + 10)

10 - x = 2x - 8

3x = 18

Thus, x = 6

Hence, option (a) is correct.

Question 50

The first term of an A.P. is p and the common difference is q, then its 10th term is

(a) q + 9p

(b) p - 9q

(c) p + 9q

(d) 2p + 9q

Solution 50

In the A.P., first term = p and common difference = q

The 10th term will be

a10 = p + (10 - 1)q

 = p + 9q

Hence, option (c) is correct.