Question

If harmonic mean of numbers $\frac{1}{2},\frac{1}{2^2},\frac{1}{2^3}...........................\frac{1}{2^{10}}$ is $\frac{\lambda }{2^{10}-1}$ then $\lambda$ is equal to

• A. $5\left(2^{10}\right)$
• B. $10\left(2^{10}\right)$
• C. $5$
• D. $10$

Answer 1

Answer: (C)

$5$

Harmonic mean of given numbers $=\frac{10}{\frac{1}{\left(1/2\right)}+\frac{1}{\left(1/2^2\right)}+\frac{1}{\left(1/2^3\right)}...+\frac{1}{\left(1/2^{10}\right)}}$

$=\frac{10}{2+2^2+2^3+...+2^{10}}$

$=\frac{10}{\frac{2\left(2^{10-1}\right)}{2-1}}[\because SumofG.P=a\frac{(r^n-1)}{r^{-1}}]$

$=\frac{5}{2^{10}-1}$

0n comparing with $\frac{\lambda }{2^{10}-1}\Rightarrow \lambda =5$

$\therefore$ option $C$ is correct.