NCERT Solutions for Class 11-science Maths Chapter 9 - Sequences and Series

Chapter 9 - Sequences and Series Exercise Ex. 9.1

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Chapter 9 - Sequences and Series Exercise Ex. 9.2

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Chapter 9 - Sequences and Series Exercise Ex. 9.3

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a, b, c, d are in G.P.

Therefore,

bc = ad      .....(1)

b2 = ac      .....(2)

c2 = bd      .....(3)

It has to be proved that,

(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2

R.H.S.

= (ab + bc + cd)2

= (ab + ad + cd)2  [Using (1)]

= [ab + d (a + c)]2

= a2b2 + 2abd (a + c) + d2(a + c)2

= a2b2 + 2a2bd + 2abcd + d2(a2 + 2ac + c2)

= a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2  [Using (1) and (2)]

= a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2

= a2b2 + a2c2 + a2d2 + b2 × b2 + b2c2 + b2d2 + c2b2 + c2 × c2 + c2d2

[Using (2) and (3) and rearranging terms]

= a2(b2 + c2 + d2) + b2(b2 + c2 + d2) + c2(b2 + c2 + d2)

= (a2 + b2 + c2) (b2 + c2 + d2)

= L.H.S

L.H.S. = R.H.S

(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2

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Chapter 9 - Sequences and Series Exercise Ex. 9.4

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Chapter 9 - Sequences and Series Exercise Misc. Ex.

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straight S subscript straight n equals straight n over 2 open square brackets straight a plus straight l close square brackets
equals 29 over 2 open square brackets 203 plus 399 close square brackets
equals 29 over 2 open square brackets 602 close square brackets
equals 29 cross times 301
equals 8729
Hence comma space the space sum space all space numbers space between space 200 space and space 400 space which space are space divisible space by space 7 space is space 8729.

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