QJEEM18Ma83db: JEE Main 2018 Mathematics - Class 11 JEEM18Ma83db Solution: If x 1 , x 2 , . . ., x n and , , . . . , are two A.P..s such…

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