Question

8.The nth term of a GP is 128 and the sum of its n terms is 225. If it's common ratio is 2, then it's first term is​

Answer 1

Correct Question:

The nth term of a GP is 128 and the sum of it's n terms is 255. If it's common difference is 2, then it's first term is?

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Answer:

$\sf{The \ first \ term \ of \ a \ GP \ is \ 1.}$

Given:

$\sf{\leadsto{The \ n^{th} \ term \ of \ a \ GP \ is \ 128.}}$

$\sf{\leadsto{The \ sum \ of \ it's \ n \ terms \ is \ 255.}}$

$\sf{\leadsto{It's \ common \ ratio \ is \ 2.}}$

To find:

$\sf{The \ first \ term \ of \ a \ GP.}$

Solution:

$\boxed{\sf{a_{n}=ar^{n-1}}}$

$\sf{According \ to \ the \ first \ condition}$

$\sf{128=a\times2^{n-1}}$

$\sf{\therefore{128=a\times\dfrac{2^{n}}{2}}}$

$\sf{\therefore{a=\dfrac{256}{2^{n}}...(1)}}$

$\boxed{\sf{S_{n}=a\times\dfrac{(r^{n}-1)}{(r-1)}}}$

$\sf{\therefore{255=a\times\dfrac{(2^{n}-1)}{(2-1)}}}$

$\sf{\therefore{a=\dfrac{255}{2^{n}-1}...(2)}}$

$\sf{from \ (1) \ and \ (2) \ we \ get,}$

$\sf{\dfrac{256}{2^{n}}=\dfrac{255}{2^{n}-1}}$

$\sf{Substitute \ 2^{n}=m}$

$\sf{\therefore{\dfrac{256}{m}=\dfrac{255}{m-1}}}$

$\sf{\therefore{256(m-1)=255m}}$

$\sf{\therefore{256m-256=255m}}$

$\sf{\therefore{256m-255m=256}}$

$\sf{\therefore{m=256}}$

$\sf{Resubstituting \ m=2^{n}}$

$\sf{\therefore{2^{n}=256}}$

$\sf{\therefore{2^{n}=2^{8}}}$

$\sf{\therefore{n=8}}$

$\sf{Substitute \ n=8 \ in \ equation \ (1), \ we \ get}$

$\sf{a=\dfrac{256}{2^{8}}}$

$\sf{\therefore{a=\dfrac{256}{256}}}$

$\sf{\therefore{a=1}}$

$\sf\purple{\tt{\therefore{The \ first \ term \ of \ a \ GP \ is \ 1.}}}$