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

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.1

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

a_{n} = n^{2} - n + 1

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

a_{n} = (n - 1) (2 - n) (3 + n); a_{1}, a_{2}, a_{3}

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

a_{n} = (-1)^{n} n; a_{3}, a_{5}, a_{8}

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.2

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

a_{n} = 2n - 1

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

a_{n} = 3n^{2} + 5

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

a_{n} = 1 + n + n^{2}

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.3

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

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.

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.

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.

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}

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.4

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

The 26^{th}, 11^{th} and last term of an A.P. are 0, 3 and , respectively. Find the common difference and the number of terms.

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

3, 8, 13, ..., 253

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

We have:

Last term (*l*) = 253

Common difference (*d*) = 8 - 3 = 5

Therefore,

12^{th} term from end

= *l* - (*n* - 1)*d*

= 253 - (12 - 1) (5)

= 253 - 55

= 198

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

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 15^{th} term.

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

Thus, n^{th} term is given by

a_{n} = a + (n - 1)d

a_{n} = 3 + (n - 1)4

a_{n} = 3 + 4n - 4

a_{n} = 4n - 1

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

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

t_{n} = 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.

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

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

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

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

t_{24} = a + (24 - 1)d = a + 23d

t_{10} = a + (10 - 1)d = a + 9d

t_{72} = a + (72 - 1)d = a + 71d

t_{15} = a + (15 - 1)d = a + 14d

t_{24} = 2t_{10}

⇒ a + 23d = 2(a + 9d)

⇒ a + 23d = 2a + 18d

⇒ 23d - 18d = 2a - a

∴ 5d = a

t_{72} = a + 71d

= 5d + 71d

= 76d

= 20d + 56d

= 4 × 5d + 4 × 14d

= 4(5d + 14d)

= 4(a + 14d)

= 4t_{15}

∴t_{72} = 4t_{15}

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

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.

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

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

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

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

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

a = 40

d = 37 - 40 = - 3

Let the n^{th} term of the series be 0.

t_{n} = a + (n - 1)d

⇒ 0 = 40 + (n - 1)( - 3)

⇒ 0 = 40 - 3(n - 1)

⇒ 3(n - 1) = 40

∴ No term of the series is 0.

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

Given A.P. is 213, 205, 197, …, 37.

Here, first term = a = 213

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

a_{n} = 37

n^{th} term of an A.P. is given by

a_{n} = 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 12^{th }term.

⇒ a_{12} = 213 + (12 - 1)(-8)

= 213 + (11)(-8)

= 213 - 88

= 125

Hence, the middle term is 125.

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

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

Then, we have

a_{5} = 31 and a_{25} = a_{5} + 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, …..

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

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

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

For the A.P.: -3, -7, -11, …, can we find a_{30} - a_{20} without actually finding a_{30} and a_{20}? Give reason for your answer.

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 10^{th} terms is the same as the difference between their 21^{st} terms, which is the same as the difference between any two corresponding terms. Why?

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.5

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.

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

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

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.

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.6

Find the sum of

The first 15 multiples of 8.

Multiples of 8 are8,16,24,…

Now,

n=15, a=8, d=8

Find the sum of

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

a) divisible by 3 3,6,9,… Now, n=40, a=3, d=3

b) divisible by 5 5,10,15,… Now, n=40, a=5, d=5

c)divisible by 6 6,12,18,… Now, n=40, a=3, d=6

Find the sum of

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

Three-digit numbers divisible by 13 are 104,117,
130,…988.
Now,
a=104, *l*=988

Find the sum of

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

Three-digit numbers which are multiples
of 11 are 110,121, 132,…990.
Now,
a=110, *l*=990

Find the sum of

All 2 - digit natural numbers divisible by 4.

Two-digit numbers divisible by 4 are 12,16,…96.
Now,
a=12, *l*=96

Let there be an A.P. with first term 'a',
common difference 'd'. If a_{n} denotes its
n^{th} term and S_{n} the sum of
first n terms, find

S_{22},
if d = 22 and a_{22} = 149

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

Find the sum of the first 15 terms of each of the following sequences having *n*^{th} term as* y _{n}* = 9 - 5

*n*

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?

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

Remark* - Question modified.

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

Find the sum of

(i) the first 15 multiples of 8

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

(iii) all 3 - digit natural numbers which are divisible by 13.

(iv) all 3 - digit natural numbers, which are multiples of 11.

Find the sum:

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

t_{n} = a + (n - 1)d

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.

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

t_{n} = a + (n - 1)d

⇒ 49 = 7 + (n - 1)d

⇒ 42 = (n - 1)d…..(i)

∴ 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

Common difference, d =3

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.

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

t_{n} = a + (n - 1)d

⇒ 45 = 5 + (n - 1)d

⇒ 40 = (n - 1)d…..(i)

∴ 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

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

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

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

t_{n} = a + (n - 1)d

t_{10} = a + (10 - 1)d

⇒ 21 = a + 9d……(i)

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

t_{n} = a + (n - 1)d

= 3 + (n - 1)2

= 3 + 2n - 2

∴ t_{n}= 2n + 1

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 28^{th} term of this A.P.

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

63 = 7[a + 3d]

9 = a + 3d……….(i)

Sum of the next 7 terms = 161

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

224 = 7[2a + 13d]

32 = 2a + 13d………..(ii)

Solving (i) and (ii), we get

d = 2, a = 3

28^{th} term of the A.P., t_{28} = a + (28 - 1)d

= 3 + 27 × 2

= 3 + 54

= 57

∴ The 28^{th} of the A.P. is 57.

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

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

26 × 2 = [2a + (7 - 1)d]

52 = 2a + 6d

26 = a + 3d……..(i)

From (i) and (ii),

⇒ 13d = 104

∴d = 8

From (i), a = 2

The A.P. is 2, 10, 18, 26,…….

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

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

t_{n} = - 4n + 15

t_{1} = - 4 × 1 + 15 = 11

t_{2} = - 4 × 2 + 15 = 7

t_{3} = - 4 × 3 + 15 = 3

Common Difference, d = 7 - 11 = -4

= 10 × (-54)

= - 540

*Note: Answer given in the book is incorrect.

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

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.

Let the number of the terms be 'n'.

Common Difference, d = - 9 + 12 = 3

t_{n} = 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, n_{1} = 12

Sum of all the terms,

The sum of the terms = 66

The sum of the first n terms of an A.P. is 3n^{2} + 6n. Find the nth term of this A.P.

Sum of n terms of the A.P., S_{n} = 3n^{2} + 6n

S_{1} = 3 × 1^{2} + 6 × 1 = 9 = t_{1} ……(i)

S_{2} = 3 × 2^{2} + 6 × 2 = 24 = t_{1} + t_{2} …….(ii)

S_{3} = 3 × 3^{2} + 6 × 3 = 45 = t_{1} + t_{2} + t_{3} ……..(iii)

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

t_{1 }= 9, t_{2} = 15, t_{3} = 21

Common difference, d = 15 - 9 = 6

n^{th} of the AP, t_{n} = a + (n - 1)d

= 9 + (n - 1) 6

= 9 + 6n - 6

= 6n + 3

Thus, the n^{th} term of the given A.P. = 6n + 3

The sum of the first n terms of an A.P. is 5n - n^{2}. Find the n^{th} term of this A.P.

S_{n} = 5n - n^{2}

S_{1} = 5 × 1 - 1^{2} = 4 = t_{1}………..(i)

S_{2} = 5 × 2 - 2^{2} = 6 = t_{1} +t_{2}………..(ii)

S_{3} = 5 × 3 - 3^{2} = 6 = t_{1 }+ t_{2 }+ t_{3}……….(iii)

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

t_{1} = 4, t_{2} = 2, t_{3} = 0

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

t_{n} = a + (n - 1)d

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

= 4 - 2n + 2

= 6 - 2n

The sum of the first n terms of an A.P. is 4n^{2} + 2n. find the n^{th} term of this A.P.

S_{n} = 4n^{2} + 2n

S_{1} = 4 × 1^{2} + 2 × 1 = 6 = t_{1}………….(i)

S_{2} = 4 × 2^{2} + 2 × 2 = 20 = t_{1} + t_{2}……….(ii)

S_{3} = 4 × 3^{2} + 2 × 3 = 42 = t_{1} + t_{2} + t_{3}………..(iii)

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

t_{1} = 6, t_{2} = 14, t_{3} = 22

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

t_{n} = a + (n - 1)d

t_{n} = 6 + (n - 1)8

= 6 + 8n - 8

= 8n - 2

*Note: Answer given in the book is incorrect.

The sum of first n terms of an A.P. is 3n^{2} + 4n. find the 25^{th} term of this A.P.

Sum of n terms of the A.P., S_{n} = 3n^{2} + 4n

S_{1} = 3 × 1^{2} + 4 × 1 = 7 = t_{1}………(i)

S_{2} = 3 × 2^{2} + 4 × 2 = 20 = t_{1} + t_{2}…….(ii)

S_{3} = 3 × 3^{2} + 4 × 3 = 39 = t_{1} + t_{2} + t_{3} …….(iii)

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

t_{1} = 7, t_{2} = 13, t_{3} = 19

Common difference, d = 13 - 7 = 6

25^{th} of the term of this A.P., t_{25} = 7 + (25 - 1)6

= 7 + 144 = 151

∴The 25^{th} term of the A.P. is 151.

The sum of first n terms of an A.P. is 5n^{2} + 3n. If its m^{th} term is 168, find the value of m. Also, find the 20^{th} term of this A.P.

Sum of the terms, S_{n} = 5n^{2} + 3n

S_{1} = 5 × 1^{2} + 3 × 1 = 8 = t_{1}………..(i)

S_{2} = 5 × 2^{2} + 3 × 2 = 26 = t_{1} + t_{2}…………..(ii)

S_{3} = 5 × 3^{2} + 3 × 3 = 54 = t_{1} + t_{2} + t_{3}…………(iii)

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

t_{1} = 8, t_{2} = 18, t_{3} = 28

Common difference, d = 18 - 8 = 10

t_{m} = 168

⇒ a + (m - 1)d = 168

⇒ 8 + (m - 1)×10 = 168

⇒ (m - 1) × 10 = 160

⇒ m - 1 = 16

∴m = 17

t_{20} = a + (20 - 1)d

= 8 + 19 × 10

= 8 + 190

= 198

The sum of first q terms of an A.P. is 63q - 3q^{2}. If its pth term -60, find the value of p. Also, find the 11^{th} term of this A.P.

Remark* - Question modified.

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

Sum of the first 'q' terms, S_{q} = 63q - 3q^{2}

S_{1} = 63 × 1 - 3 × 1^{2} = 60 = t_{1}……..(i)

S_{2} = 63 × 2 - 3 × 2^{2} = 114 = t_{1} + t_{2}…..(ii)

S_{3} = 63 × 3 - 3 × 3^{2} = 162 = t_{1} + t_{2 }+ t_{3} .....(iii)

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

t_{1} = 60

t_{2} = 54

t_{3} = 48

Common difference, d = 54 - 60 = - 6

t_{p} = a + (p - 1)d

⇒ -60 = 60 + (p - 1)( - 6)

⇒ - 120 = - 6(p - 1)

⇒ p - 1 = 20

∴p = 21

t_{11} = 60 + (11 - 1)( - 6)

=60 + 10( - 6)

= 60 - 60

= 0

The 11^{th} term of the A.P. is 0.

The sum of first m terms of an A.P. is 4m^{2} - m. If its nth term is 107, find the value of n. Also, find the 21^{st} term of this A.P.

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

Sum of m terms of the A.P., S_{m} = 4m^{2} - m

S_{1} = 4 × 1^{2} - 1 = 3 = t_{1} …….(i)

S_{2} = 4 × 2^{2} - 2 = 14 = t_{1} + t_{2}……..(ii)

S_{3} = 4 × 3^{2} - 3 = 33 = t_{1 }+ t_{2} + t_{3} …….(iii)

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

t_{1} = 3, t_{2} = 11, t_{3} = 19

Common difference, d = 11 - 3 = 8

t_{n} = 107

⇒ a + (n - 1)d = 107

⇒ 3 + (n - 1)8 = 107

⇒ 8(n - 1) = 104

⇒ n - 1 = 13

∴n = 14

t_{21} = 3 + (21 - 1)8 = 3 + 160 = 163

If the sum of first n terms of an A.P. is then find its n^{th} term. Hence write its 20^{th} term.

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

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

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

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

Let there be an A.P. with first term 'a', common difference 'd'. If a_{n} denotes its n^{th} term and S_{n} the sum of first n terms, find

n and a_{n}, if a = 2, d = 8 and S_{n} = 90.

Let there be an A.P. with first term 'a', common difference 'd'. If a_{n} denotes its n^{th} term and S_{n} the sum of first n terms, find k, if S_{n} = 3n^{2} + 5n and a_{k} = 164.

If S_{n} denotes the sum of first n terms of an A.P., prove that S_{12} = 3(S_{8} - S_{4}).

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?

The sums of first n terms of three A.P.s are S_{1}, S_{2} and S_{3}. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that S_{1} + S_{3} = 2S_{2}.

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?

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.

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,

Number of trees planted by the students = 312

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?

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

= 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.

If S_{n} denotes the sum of the first n terms of an A.P., prove that S_{30} = 3(S_{20} - S_{10}).

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

R.H.S.

= 3(S_{20} - S_{10})

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

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

= 3(10a + 145d)

= 3 × 5(2a + 29d)

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

= S_{30}

= L.H.S.

Solve the equation

(-4) + (-1) + 2 + 5 + …. + x = 437.

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

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.

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?

### Other Chapters for CBSE Class 10 Mathematics

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Pairs of Linear Equations in Two Variables Chapter 4- Quadratic Equations Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- Trigonometric Identities Chapter 12- Heights and Distances Chapter 13- Areas Related to Circles Chapter 14- Surface Areas and Volumes Chapter 15- Statistics Chapter 16- Probability### RD SHARMA Solutions for CBSE Class 10 Subjects

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