Question

If $9$ AM’s and HM’s are inserted between $2$ and $3$ and if the harmonic mean $H$ is corresponding to arithmetic mean $A$, then $A+\frac{6}{H}$ is equal to

• A. $1$
• B. $3$
• C. $5$
• D. $6$

Answer 1

The correct option is C

$5$

Explanation of the correct option :

Step 1. Assume the variables:

Let $A_1,A_2\dots A_9$ are the AM’s and $H_1,H_2,\dots H_9$ are the HM’s between $a$ and $b$.

Given, $9$ AM’s and HM’s are inserted between $2$ and $3$.

Total number of terms$=9+2$

$=11$

Step 2. Consider, the Arithmetic progression:

Let it be $2,A_1,A_2,.....A_9,3$

$a_{11}=a+10d$

$\Rightarrow$$2+10d=3$

$\Rightarrow$ $10d=1$

$\Rightarrow$ $d=\frac{1}{10}$

$A_1=2+d$

$\Rightarrow$$A_n=2+\frac{n}{10}$

Step 3. Consider the Harmonic progression:

Let it be $2,H_1,H_2,\dots .H_9,3$

So $\frac{1}{2},\frac{1}{H_1},\frac{1}{H_2}\dots ..\frac{1}{H_9},\frac{1}{3}$ are in AP

$a_{11}=a+10D$

$\Rightarrow$$\frac{1}{2}+10D=\frac{1}{3}$

$\Rightarrow$ $10D=-\frac{1}{6}$

$\Rightarrow$ $D=-\frac{1}{60}$

$\therefore$$\frac{1}{H_1}=\frac{1}{2}-\frac{1}{60}$

so the nth term is $\frac{1}{H_n}=\frac{1}{2}–\frac{n}{60}$

$\Rightarrow$ $\frac{6}{H_n}=3–\frac{n}{10}$

$\therefore$ $\begin{array}{rcl}{A}_n+\frac{6}{H_n}& =& 2+\frac{n}{10}+3–\frac{n}{10}\end{array}$

$\begin{array}{rcl}& =& \mathbf{}\mathbf{5}\end{array}$

Hence, option C. is the answer.