Question

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

Answer 1

The corresponding $T_3$ and $T_7$ of AP are, $T_3=7,T_7=5$
$d=\frac{T_p-T_q}{p-q}=\frac{T_7-T_3}{7-3}=\frac{5-7}{4}=\frac{-2}{4}=\frac{-1}{2}$
$\therefore d=\frac{-1}{2}$
Consider $T_3$ of the corresponding AP
$T_3=7$
$a+2d=7\Rightarrow a+2\left(\frac{-1}{2}\right)=7$
$a-1=7$
$\therefore a=8$
To find $T_{15},T_n=a+(n-1)d$
$T_{15}=8+(15-1)\left(\frac{-1}{2}\right)$
$T_{15}=8+14\left(\frac{-1}{2}\right)$
$T_{15}=8-7=1$
In A.P., $T_{15}=1$
$\Rightarrow$ Corresponding $T_{15}$ of the H.P., $T_{15}=1$