Question

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

• A. ${10.2}^{10}$
• B. $5$
• C. ${5.2}^{10}$
• D. 10

Answer 1

Answer: (B)

$5$

Given quantities are $\frac{1}{2},\frac{1}{2^2},\frac{1}{2^3},.....\frac{1}{2^{10}},$
H.M = $\frac{10}{2+2^2+2^3+..+2^{10}}$

$=\frac{10}{\frac{2(2^{10}-1)}{2-1}}$

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

$\therefore \lambda =5$