Let the A.P. $x_1,x_2,\dots,x_n$ have first term $a$ and common difference $d$. Then $x_k=a+(k-1)d.$

Also, $\left(\frac1{h_1},\frac1{h_2},\dots,\frac1{h_n}\right)$ is an A.P. Let its first term be $p$ and common difference $q$. So, $\frac1{h_k}=p+(k-1)q.$

We are given:

• $x_3=8$
• $x_8=20$
• $h_2=8$
• $h_7=20$

We need to find $x_5\cdot h_{10}$.

1. Find $x_5$

From the A.P. of $x_k$: $x_3=a+2d=8$ $x_8=a+7d=20$ Subtracting, $5d=12 \implies d=\frac{12}{5}.$ Then $x_5=x_3+2d=8+2\cdot \frac{12}{5}=8+\frac{24}{5}=\frac{64}{5}.$

2. Find $h_{10}$

Since $\left(\frac1{h_k}\right)$ is an A.P., use the given values: $h_2=8 \implies \frac1{h_2}=\frac18$ $h_7=20 \implies \frac1{h_7}=\frac1{20}$

Thus, $\frac1{h_2}=p+q=\frac18,$ $\frac1{h_7}=p+6q=\frac1{20}.$ Subtracting, $5q=\frac1{20}-\frac18=\frac{2-5}{40}=-\frac3{40}$ $q=-\frac3{200}.$

Now, \frac1{h_{10}}=\frac1{h_2}+8q=rac18+8\left(-\frac3{200}\right). Since from $h_2$ to $h_{10}$ there are $8$ steps, \frac1{h_{10}}=\frac18-\frac{24}{200}=rac18-\frac3{25}. Take LCM $200$: $\frac1{h_{10}}=\frac{25}{200}-\frac{24}{200}=\frac1{200}.$ Hence, $h_{10}=200.$

3. Compute $x_5\cdot h_{10}$

$x_5\cdot h_{10}=\frac{64}{5}\cdot 200=64\cdot 40=2560.$

4. Check options

The value is $\boxed{2560}$ So the correct option is A.

5. Comparison with stored answer

Stored correct answer: A

Our derived answer is also A. Hence, the stored answer is correct.