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

We have two arithmetic progressions. The first one is $$x_1,\,x_2,\,\ldots,x_n$$ and the second one is $$\dfrac1{h_1},\,\dfrac1{h_2},\,\ldots,\,\dfrac1{h_n}.$$ The given numerical data are $$x_3=h_2=8$$ and $$x_8=h_7=20.$$ Our aim is to find the product $$x_5\cdot h_{10}.$$

For any A.P. the general term is governed by the well-known relation

$$T_k=T_1+(k-1)d,$$

where $$T_1$$ is the first term and $$d$$ is the common difference. We apply this relation separately to the two progressions.

First progression — the $$x_k$$ sequence. Let us denote $$x_1=a$$ and its common difference by $$d.$$ Then

$$x_k=a+(k-1)d.$$

Using the given values, we substitute $$k=3$$ and $$k=8$$ one by one:

$$x_3=a+2d=8\qquad\text{(1)}$$

$$x_8=a+7d=20\qquad\text{(2)}$$

We subtract equation (1) from equation (2):

$$\bigl(a+7d\bigr)-\bigl(a+2d\bigr)=20-8\;\;\Longrightarrow\;\;5d=12,$$

hence

$$d=\frac{12}{5}=2.4.$$

Substituting $$d$$ back in equation (1) gives the first term $$a$$:

$$a+2\left(\frac{12}{5}\right)=8\;\;\Longrightarrow\;\;a=8-\frac{24}{5}=\frac{16}{5}=3.2.$$

Now we calculate $$x_5$$ by putting $$k=5$$ in the general term:

$$x_5=a+4d=\frac{16}{5}+4\left(\frac{12}{5}\right)=\frac{16}{5}+\frac{48}{5}=\frac{64}{5}=12.8.$$

Second progression — the reciprocals $$\dfrac1{h_k}$$. Let

$$\frac1{h_1}=b,$$

and let its common difference be $$e.$$ Then

$$\frac1{h_k}=b+(k-1)e.$$

The values of $$h_2$$ and $$h_7$$ are supplied, so their reciprocals are known:

$$\frac1{h_2}=\frac18,\qquad\frac1{h_7}=\frac1{20}.$$

Writing these in the general form gives two linear equations:

$$b+e=\frac18\qquad\text{(3)}$$

$$b+6e=\frac1{20}\qquad\text{(4)}$$

Subtract equation (3) from equation (4):

$$(b+6e)-(b+e)=\frac1{20}-\frac18\;\;\Longrightarrow\;\;5e=\frac1{20}-\frac18.$$

The numerical difference on the right side is

$$\frac1{20}-\frac18=\frac{1}{20}-\frac{5}{40}=\frac{2}{40}-\frac{5}{40}=-\frac{3}{40},$$

so

$$5e=-\frac{3}{40}\;\;\Longrightarrow\;\;e=-\frac{3}{200}.$$

Now we determine $$b$$ from equation (3):

$$b=\frac18-e=\frac18+\frac{3}{200}=\frac{25}{200}+\frac{3}{200}=\frac{28}{200}=\frac7{50}.$$

Next we find $$h_{10}$$. First we write its reciprocal:

$$\frac1{h_{10}}=b+9e=\frac7{50}+9\left(-\frac{3}{200}\right)=\frac7{50}-\frac{27}{200}.$$

Converting $$\dfrac7{50}$$ to the denominator $$200$$ gives $$\dfrac{28}{200}$$, hence

$$\frac1{h_{10}}=\frac{28}{200}-\frac{27}{200}=\frac1{200},$$

so

$$h_{10}=200.$$

Finally, the desired product.

$$x_5\cdot h_{10}=\left(\frac{64}{5}\right)\times200=64\times40=2560.$$

Hence, the correct answer is Option A.