Question

4 HMs are inserted between 2 and 5, the 4th term in the H.P so formed is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Let $H_1$, $H_2$, $H_3$, $H_4$ be HM's inserted

2, $H_1$, $H_2$, $H_3$, $H_4$, 5 form H.P.

Common difference of corresponding A.P. is

D = $\frac{a-b}{(n+1)ab}$

$=\frac{2-5}{5\times 10}$

= $\frac{-3}{50}$

Let $t_n$ be nth term of H.P.

$t_1$ = 2, D = $\frac{-3}{50}$

$t_n=\frac{1}{\frac{1}{t_1}+(n-1)D}=\frac{1}{\frac{1}{2}-\frac{3}{50}(n-1)}=\frac{50}{25-3(n-1)}$

$so,t_4=\frac{50}{25-3\left(3\right)}=\frac{50}{16}=\frac{25}{8}$