Question

If between $1$ and $1/31$ there are $n$ $H.M.s$ and ratio of $7$th and $(n–1)$th harmonic means is $9:5$. then value of $n$ is

• A. $12$
• B. $13$
• C. $14$
• D. $15$

Answer 1

The correct option is C

$14$

Step 1. Find the value of $n$:

Given $n$$H.M.s$ between $1and\frac{1}{31}$.

So common difference , $d=\frac{(1-{\displaystyle \frac{1}{31}})}{(\mathrm{n}+1)\left({\displaystyle \frac{1}{31}}\right)}$

$=\frac{\left({\displaystyle \frac{31-1}{31}}\right)}{\left(n+1\right)\left({\displaystyle \frac{1}{31}}\right)}$

$=\frac{30}{(\mathrm{n}+1)}$

Step 2. Find the value of ${\mathit{h}}_{\mathbf{1}}\mathbf{,}\mathbf{}{\mathit{h}}_{\mathbf{7}}\mathbf{,}\mathbf{}{\mathit{h}}_{\mathbf{n}\mathbf{-}\mathbf{1}}$,

$\begin{array}{rcl}\frac{1}{{\mathrm{h}}_1}& \mathit{=}& \mathit{1}\mathit{+}d\\ & \mathit{=}& \frac{(n+31)}{(n+1)}\end{array}$

$\begin{array}{rcl}\frac{1}{{\mathrm{h}}_7}& \mathit{=}& \mathit{1}\mathit{+}\mathit{7}d\\ & \mathit{=}& \mathit{1}\mathit{+}\mathit{7}\mathit{\times }\frac{\mathit{30}}{n\mathit{+}\mathit{1}}\\ & \mathit{=}& \frac{(n+211)}{\mathit{(}n\mathit{+}\mathit{1}\mathit{)}}\end{array}$

$\Rightarrow$$h_7=\frac{(\mathrm{n}+1)}{(\mathrm{n}+211)}$

$\begin{array}{rcl}\frac{\mathit{1}}{h_{n\mathit{-}\mathit{1}}}& \mathit{=}& \mathit{1}\mathit{+}\mathit{(}n\mathit{-}\mathit{1}\mathit{)}d\\ & \mathit{=}& \mathit{1}\mathit{+}\mathit{(}n\mathit{-}\mathit{1}\mathit{)}\mathit{\times }\frac{\mathit{30}}{(n+1)}\\ & \mathit{=}& \frac{(31n-29)}{(n+1)}\end{array}$

$\Rightarrow$$h_{\mathrm{n}-1}=\frac{(\mathrm{n}-1)}{(31\mathrm{n}-29)}$

Step 3. Divide $h_7$ by $h_{n-1}$, we get

$\frac{h_7}{h_{n-1}}=\frac{(31n-29)}{(n+211)}$

$\Rightarrow$ $\frac{9}{5}=\frac{(31n-29)}{(n+211)}$

Cross multiply

$\Rightarrow$$9(\mathrm{n}+211)=5(31\mathrm{n}-29)$

$\Rightarrow$$9n+1899=155n-145$

$\Rightarrow$$9n-155n=-145-1899$

$\Rightarrow$ $-146n=-2044$

$\Rightarrow$ $146n=2044$

$\Rightarrow$ $n=\frac{2044}{146}$

$\mathbf{\therefore }\mathit{n}\mathbf{=}\mathbf{14}$

Hence option(C) is correct..