Question

1. The 4^th term of a H.P. is $\frac{3}{5}$ and 8^th term is $\frac{1}{3}$ then its 6^th term is
(a) $\frac{1}{6}$ (b) $\frac{3}{7}$ (c) $\frac{1}{7}$ (4) $\frac{3}{5}$

Answer 1

Dear Student
$$\begin{gathered} a_4=\frac{3}{5}and{a}_8=\frac{1}{3} \\ weknowthat{a}_n=\frac{1}{a+(n-1)d} \\ so{a}_4=\frac{1}{a+3d}=\frac{3}{5}and{a}_8=\frac{1}{a+7d}=\frac{1}{3} \\ so3a+9d=5-------\left(1\right) \\ anda+7d=3-------\left(2\right) \\ \left(1\right)-\left(2\right)\times 3;weget \\ 3a+9d-3a-21d=5-9 \\ \Rightarrow -12d=-4\Rightarrow d=\frac{1}{3}(putin1) \\ 3a+9\times \frac{1}{3}=5\Rightarrow 3a=5-3 \\ \Rightarrow a=\frac{2}{3} \\ now{a}_6=\frac{1}{a+5d}=\frac{1}{\left({\displaystyle \frac{2}{3}}+5\times {\displaystyle \frac{1}{3}}\right)}=\frac{1}{\left({\displaystyle \frac{7}{3}}\right)}=\frac{3}{7} \end{gathered}$$
Regards