$x_1,x_2,\dots ,x_n$and $\frac{1}{h_1},\frac{1}{h_2},\dots ..,\frac{1}{h_n}$are two A.P.s such that $x_3=h_2=8$and $x_8=h_7=20$, then $x_5{h}_{10}$ equals

Let $d_1$and $1/d_2$be the common difference of A.P.

$x_1,x_2,\dots .,x_n\text{ and }\frac{1}{h_1},\frac{1}{h_2},\frac{1}{h_3},\dots ,\frac{1}{h_n}$ respectively.

Given,$x_3=8;x_8=20$ 

$\Rightarrow x_1+2d_1=8$               ..(i)         $x_1+7d_1=20$            (ii)

On solving (i) and (ii), we get $d_1=\frac{12}{5}\text{ and }{x}_1=\frac{16}{5}$

Similarly; $h_2=8$and $h_7=20$

$\Rightarrow \frac{1}{h_2}=\frac{1}{8}$and $\frac{1}{h_7}=\frac{1}{20}$

$\Rightarrow \frac{1}{h_1}+\frac{1}{d_2}=\frac{1}{8}\text{...(iii)}$ and $\frac{1}{h_1}+\frac{6}{d_2}=\frac{1}{20}$            ..(iv)

On solving (iii) and (iv), we get $\frac{1}{d_2}=\frac{-3}{200},\frac{1}{h_1}=\frac{28}{200}$

Now, $x_5=x_1+4d_1=\frac{16}{5}+\frac{48}{5}=\frac{64}{5}$

Also, $\frac{1}{h_{10}}=\frac{1}{h_1}+\frac{9}{d_2}=\frac{28}{200}-\frac{27}{200}=\mid \frac{1}{200}$

$\therefore x_5\cdot h_{10}=\frac{64}{5}\times 200=2560$