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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Sequences and Series

NDA 2023 · Mathematics · Sequences and Series

If H is the harmonic mean of numbers 1,2.2^2, 2^3, , 2 n^ -1 , then...

If HH is the harmonic mean of numbers 1,2.22,23,…1,2.2^2, 2^3, \ldots, 2n−12 n^{-1}, then what is n/Hn / H equal to?

Options

  1. A2−12n+12-\frac{1}{2^{n+1}}
  2. B2−12n−12-\frac{1}{2^{n-1}}
  3. C2+12n−12+\frac{1}{2^{n-1}}
  4. D2−12n2-\frac{1}{2^n}

Correct answer

B. 2−12n−12-\frac{1}{2^{n-1}}

Step-by-step solution

∵n1a1+1a2+……+1an=HnH=1a1+1a2+……+1an=1+12+122+………+12n−1=1((12)n−1)(12−1)=2(1−12n)∴nH=2−12n−1\because \frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\ldots \ldots+\frac{1}{a_n}}=H\frac{n}{H}=\frac{1}{a_1}+\frac{1}{a_2}+\ldots \ldots+\frac{1}{a_n}=1+\frac{1}{2}+\frac{1}{2^2}+\ldots \ldots \ldots+\frac{1}{2^{n-1}}=\frac{1\left(\left(\frac{1}{2}\right)^n-1\right)}{\left(\frac{1}{2}-1\right)}=2\left(1-\frac{1}{2^n}\right)\therefore \quad \frac{n}{H}=2-\frac{1}{2^{n-1}}
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