Question

If $n$ is the total number of harmonic mean between $a$ and $b$, then $m^{th}$ harmonic mean between $a$ and $b$ is

• A. $\frac{(n+1)ab}{(n+1)}+n(a-b)$
• B. $\frac{(n+1)ab}{(n+1)b+m(a-b)}$
• C. $(n+1)b+m(a-b)$
• D. $(n+1)ba$

Answer 1

The correct option is B $\frac{(n+1)ab}{(n+1)b+m(a-b)}$

The series $a,H_1,H_2,H_3.....H_n,b$ is in harmonic progression.

$\Rightarrow \frac{1}{a},\frac{1}{H_1},\frac{1}{H_2},\frac{1}{H_3}.............\frac{1}{H_n},\frac{1}{b}$ are in A.P.

Total terms in A.P $=n+2$

General term of $A.P.$ is $a_t=a+(t-1)d$

$\Rightarrow a_{n+2}=a+(n+2-1)d\Rightarrow a_{n+2}=a+(n+1)d\Rightarrow \frac{1}{b}=\frac{1}{a}+(n+1)d\Rightarrow d=\frac{(a-b)}{ab(n+1)}$

Now term corresponding to $m^{th}$ H.M is $(m+1)$th term in A.P

$\Rightarrow a_{m+1}=a+(m+1-1)d\Rightarrow a_{m+1}=a+md\Rightarrow a_{m+1}=\frac{1}{a}+m\left(\frac{(a-b)}{ab(n+1)}\right)\Rightarrow \frac{1}{H_m}=\frac{b(n+1)+m(a-b)}{ab(n+1)}\Rightarrow H_m=\frac{ab(n+1)}{b(n+1)+m(a-b)}$