Assume d₁ is the common difference of A.P x₁,x₂ ..... x_n

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