How many terms of the AP : 9, 17, 25,……. must be taken to give a sum of 636?

Given,

A.P. : 9, 17, 25,…….

First term (a) = 9 and common difference (d) = 17 - 9 = 8.

Let sum of n terms of the above A.P. be 636.

∴ S_n = 636

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow 636 = \dfrac{n}{2}[2 \times 9 + (n - 1) \times 8] \\[1em] \Rightarrow 636 = \dfrac{n}{2}[18 + (n - 1) \times 8] \\[1em]\Rightarrow 636 = \dfrac{n}{2}[18 + 8n - 8] \\[1em] \Rightarrow 636 = \dfrac{n}{2}[8n + 10] \\[1em] \Rightarrow 636 = \dfrac{8n^2 + 10n}{2} \\[1em] \Rightarrow 8n^2 + 10n = 1272 \\[1em] \Rightarrow 2(4n^2 + 5n) = 1272 \\[1em] \Rightarrow 4n^2 + 5n = \dfrac{1272}{2} \\[1em] \Rightarrow 4n^2 + 5n = 636 \\[1em] \Rightarrow 4n^2 + 5n - 636 = 0$

Comparing equation,

4n² + 5n - 636 = 0 with ax² + bx + c = 0, we get :

a = 4, b = 5 and c = -636.

By formula,

n = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$\Rightarrow n = \dfrac{-5 \pm \sqrt{5^2 - 4 \times 4 \times -636}}{2 \times 4} \\[1em] = \dfrac{-5 \pm \sqrt{25 + 10176}}{8} \\[1em] = \dfrac{-5 \pm \sqrt{10201}}{8} \\[1em] = \dfrac{-5 \pm 101}{8} \\[1em] = \dfrac{-5 + 101}{8}, \dfrac{-5 - 101}{8} \\[1em] = \dfrac{96}{8}, -\dfrac{106}{8} \\[1em] = 12, -\dfrac{53}{4}$

Since, no. of term cannot be negative.

∴ n = 12.

Hence, sum of 12 terms of A.P. = 636.