Sequence And Series Question 328
Question: Let $ A_{n} $ be the sum of the first n terms of the geometric series $ 704+\frac{704}{2}+\frac{704}{4}+\frac{704}{8}+…… $ and $ B_{n} $ be the sum of the first n terms of the geometric series $ 1984+\frac{1984}{2}+\frac{1984}{4}+\frac{1984}{8}+…… $ If $ A_{n}=B_{n}, $ then the value of n is (where $ n\in N $ ).
Options:
4
5
6
7
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ A_{n}=704+\frac{704}{2}+\frac{704}{4}+…. $ to n terms $ =\frac{704( 1-{{( \frac{1}{2} )}^{n}} )}{1-\frac{1}{2}}=704\times 2( 1-{{( \frac{1}{2} )}^{n}} ) $ $ B_{n}=1984-\frac{1984}{2}+\frac{1984}{4}…. $ to n terms $ =\frac{1984( 1-{{( \frac{-1}{2} )}^{n}} )}{1-( \frac{-1}{2} )}=1984\times \frac{2}{3}( 1-{{( \frac{-1}{2} )}^{n}} ) $ Now, $ A_{n}=B_{n}\Rightarrow 704\times 2( 1-{{( \frac{1}{2} )}^{n}} ) $ $ =1984\times \frac{2}{3}\times ( 1-{{( \frac{-1}{2} )}^{n}} ) $
$ \Rightarrow 33-31=33{{( \frac{1}{2} )}^{n}}-31{{( \frac{-1}{2} )}^{n}} $
$ \Rightarrow {2^{n+1}}=33-31{{(-1)}^{n}}\Rightarrow n=4 $
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