Question

If $H_1,H_2,\dots ,H_{.20}$ be $20$ harmonic means between $2$ and $3,$ then $\frac{H_1+2}{H_1-2}+\frac{H_{20}+3}{H_{20}-3}=?$

• A. $20$
• B. $21$
• C. $40$
• D. $38$

Answer 1

The correct option is A $20$

$\begin{array}{c}\text{Given that,}\\ H_1,H_2,H_3\dots H_{20}\text{are}20\text{HM between}2\text{and}3\\ \text{we know that}\\ \begin{array}{cc}d& =\frac{\frac{1}{b}-\frac{1}{a}}{n+1}=\frac{\frac{1}{3}-\frac{1}{2}}{20+1}=\frac{-1}{126}\\ \frac{1}{H_1}& =\frac{1}{a}+d=\frac{1}{2}-\frac{1}{126}=\frac{62}{126}\\ \Rightarrow & H_1=\frac{126}{62}\end{array}\end{array}$

$\begin{array}{c}\text{Similarly,}\\ \begin{array}{c}\frac{1}{H_{20}}=\frac{1}{a}+20d=\frac{1}{2}-\frac{20}{126}=\frac{43}{126}\\ \Rightarrow H_{20}=\frac{126}{43}\end{array}\\ \therefore \begin{array}{cc}\frac{H_1+2}{H_2-2}+\frac{H_{20}+3}{H_{20}-3}& =\frac{\frac{126}{62}+2}{\frac{126}{62}-2}+\frac{\frac{126}{43}+3}{\frac{126}{43}-3}\\ =& 125+(-85)=40\end{array}\end{array}$

Ans C