Question

If $a_1,a_2,a_3,.....,a_{10}$ be in A.P and $h_1,h_2,h_3,.....,h_{10}$ be in H.P. If $a_1=h_1=2$ and $a_{10}=h_{10}=3$, then the value of $a_4{h}_7$ is ....................

Answer 1

The common difference of the A.P. is $d=\frac{a_{10}-a_1}{9}=\frac{1}{9}$
Thus, $a_4=a_1+3d=\frac{7}{3}$
Similarly for the H.P., $h_1=2$ and $h_{10}=3$
So, $\frac{1}{h_1},\frac{1}{h_2},..$ being in A.P., we have the common difference to be $\frac{\frac{1}{h_{1}0}-\frac{1}{h_1}}{9}=\frac{-1}{54}$
Implying that $\frac{1}{h_7}=\frac{1}{2}-6\frac{1}{54}=\frac{7}{18}$
$\Rightarrow h_7=\frac{18}{7}$
$\Rightarrow a_4{h}_7=\frac{7}{3}\times \frac{18}{7}=6$