Question

In a harmonic progression,
$T_3=\frac{1}{7}$ and $T_7=\frac{1}{5}$. Find $T_{15}$.

Answer 1

Given :

$T_3=\frac{1}{7}$ and $T_7=\frac{1}{5}$.

The $n^{th}$ term $T_n$ of H.P in terms of A.P is given by $T_n=\frac{1}{a+(n-1)d}$,

where $\frac{1}{a}$ is first term and $d$ is common difference of H.P.

Take $n=3$,

$\therefore$ $T_3=\frac{1}{a+2d}=\frac{1}{7}$

$⟹$ $a+2d=7$ ................ $\left(1\right)$

Also $T_7=\frac{1}{a+6d}=\frac{1}{5}$

$⟹$ $a+6d=5$ ................ $\left(2\right)$

Solving equations $\left(1\right)$ and $\left(2\right)$ simultaneously we get,

$a=8$ and $d=-\frac{1}{2}$

$\therefore$ $T_{15}=\frac{1}{a+14d}=\frac{1}{8+14(-\frac{1}{2})}=\frac{1}{8-7}$

$\therefore$ $T_{15}=1$