Official solution
Assume d1 is the common difference of A.P x1,x2 ..... xn
Given x3 = 8 and x8 = 20
\[ \therefore \] x1 + 2d1 = 8 ..... (i)
and x1 + 7d1 = 20 ..... (ii)
Solving (i) and (ii) we get x1 = \[16 \over {15}\] and d1 = \[12 \over {5}\]
Now let \[1 \over d_2\] is the common difference of A.P \[1 \over h_1\], \[1 \over h_2\] ..... \[1 \over h_n\]
Given that,
h2 = 8 and h7 = 20
\[ \therefore \] \[1 \over h_2\] = \[1 \over 8\]
\[ \Rightarrow \] \[1 \over h_1\] + \[1 \over d_2\] = \[1 \over 8\] .... (iii)
and \[1 \over h_7\] = \[1 \over 20\]
\[ \Rightarrow \] \[1 \over h_1\] + \[6 \over d_2\] = \[1 \over 20\] ... (iv)
Solving (iii) and (iv) we get
\[1 \over h_1\] = \[28 \over 200\] and \[1 \over d_2\] = \[- {3 \over 200}\]
So, x5 = x1 + 4d1
= \[16 \over 5\] + \[48 \over 5\]= \[64 \over 5\] and
\[1 \over h_{10}\] = \[1 \over h_1\] + \[9 \over d_2\]
= \[28 \over 200\] - \[27 \over 200\] = \[1\over 200\]
\[ \therefore \] x5 \[\times\] h10 = \[{64 \over 5} \times 200\] = 2560