Sequence And Series Question 342
Question: What is the greatest value of the positive integer n satisfying the condition $ 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+….+\frac{1}{{2^{n-1}}}<2-\frac{1}{1000} $ ?
Options:
A) 8
B) 9
C) 10
D) 11
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+……….+\frac{1}{{2^{n-1}}}<2-\frac{1}{1000} $ LHS of given inequality is in G.P.
$ \therefore \frac{1-\frac{1}{2^{n}}}{1-\frac{1}{2}}<2-\frac{1}{1000} $
$ \Rightarrow 2-\frac{1}{{2^{n-1}}}<2-\frac{1}{100} $
$ \Rightarrow {2^{n-1}}<1000 $ Now, $ {{(2)}^{9}}=512\And {{(2)}^{10}}=1024 $
$ \therefore n-1=9\Rightarrow n=10 $ .
BETA
