Question

In the H.P. $\frac{1}{5},\frac{1}{3},1,-1,$ ........ find $T_{10}$

Answer 1

Reciprocals of the given HP are in A.P.
$\therefore 5,3,1,-1,$. ............. are in A.P.
$a=5,d=T_2-T_1=3-5=-2$
$T_{10}=?$
$T_n=a+(n-1)d$
$T_{10}=5+(10-1)(-2)=5+9(-2)=5-18=-13$
$\therefore T_{10}$ of A.P. is - 13 and $T_{10}$ of the corresponding H.P. is $\frac{-1}{13}$