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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Sequences and Series

NDA 2016 · Mathematics · Sequences and Series

What is the greatest value of the positive integer n satisfying the...

What is the greatest value of the positive integer n satisfying the condition 1+12+14+18+…..+12n−1<2−11000?1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots..+\frac{1}{2^{\mathrm{n}-1}} < 2-\frac{1}{1000} ?

Options

  1. A88
  2. B99
  3. C1010
  4. D1111

Correct answer

C. 1010

Step-by-step solution

1+12+14+18+……………+12n−1<2−110001+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \ldots \ldots \ldots \ldots+\frac{1}{2^{n-1}} < 2-\frac{1}{1000} LHS of given inequality is in G.P. ∴1−12n1−12<2−11000⇒2−12n−1<2−11000⇒2n−1<1000\begin{aligned} & \therefore \quad \frac{1-\frac{1}{2^{\mathrm{n}}}}{1-\frac{1}{2}} < 2-\frac{1}{1000} \\ & \Rightarrow \quad 2-\frac{1}{2^{\mathrm{n}-1}} < 2-\frac{1}{1000} \\ & \Rightarrow \quad 2^{\mathrm{n}-1} < 1000 \end{aligned} Now, (2)9=512(2)10=1024∴n−1=9⇒n=10.\begin{aligned} & \text{ Now, }(2)^9=512 &(2)^{10}=1024 \\ & \therefore \quad \mathrm{n}-1=9 \\ & \Rightarrow \quad \mathrm{n}=10 .\end{aligned}$
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