JEE Main 2018 — Sequences And Series Question with Solution

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 = $$\frac{16}{15}$$ and d₁ = $$\frac{12}{5}$$

Now let $$\frac{1}{d_2}$$ is the common difference of A.P $$\frac{1}{h_1}$$, $$\frac{1}{h_2}$$ ..... $$\frac{1}{h_n}$$

Given that,
h₂ = 8 and h₇ = 20

$$\therefore$$ $$\frac{1}{h_2}$$ = $$\frac{1}{8}$$

$$\Rightarrow$$ $$\frac{1}{h_1}$$ + $$\frac{1}{d_2}$$ = $$\frac{1}{8}$$ .... (iii)

and $$\frac{1}{h_7}$$ = $$\frac{1}{20}$$

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

Solving (iii) and (iv) we get
$$\frac{1}{h_1}$$ = $$\frac{28}{200}$$ and $$\frac{1}{d_2}$$ = $$-\frac{3}{200}$$
So, x₅ = x₁ + 4d₁
= $$\frac{16}{5}$$ + $$\frac{48}{5}$$= $$\frac{64}{5}$$ and
$$\frac{1}{h_{10}}$$ = $$\frac{1}{h_1}$$ + $$\frac{9}{d_2}$$
= $$\frac{28}{200}$$ - $$\frac{27}{200}$$ = $$\frac{1}{200}$$

$$\therefore$$ x₅ $$\times$$ h₁₀ = $$\frac{64}{5}\times 200$$ = 2560