Question

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

• A. ${10.2}^{10}$
• B. $5$
• C. ${5.2}^{10}$
• D. none of these

Answer 1

The correct option is B $5$

Given $\{\frac{1}{2},\frac{1}{2^2},\frac{1}{2^3},\frac{1}{2^4},...\dots ,\frac{1}{2^{10}}\}$

Harmonic mean of above data

$=\frac{10}{\frac{1}{\frac{1}{2}}+\frac{1}{\frac{1}{2^2}}+\frac{1}{\frac{1}{2^3}}+...\dots +\frac{1}{\frac{1}{2^{10}}}}$

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

$=\frac{10}{\frac{2(2^{10}-1)}{2-1}}$ [sun of first n terms of a G.P]

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

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

Comparing with $\frac{\lambda }{2^{10}-1}$

$\Rightarrow \lambda =5$.