Question

How many terms of the AP: 9, 17, 25, ... must be added to get a sum of 636?___

Answer 1

First term = a = 9

Common difference = d = 17 -9 = 8

$S_n$=636

Applying formula, $S_n$=$\frac{n}{2}$(2a+ (n−1) d) to find sum of n terms of AP, we get

636=$\frac{n}{2}$(18+ (n−1) (8))

$\Rightarrow$1272=n (18+8n−8)

$\Rightarrow$1272=18n+8$n^2$−8n

$\Rightarrow$ 8$n^2$+10n−1272=0

$\Rightarrow$4$n^2$+5n−636=0

It is a quadratic equation. We can solve it using different methods. Let’s solve it using quadratic formula.

Comparing equation 4$n^2$+5n−636=0 with general form a$n^2$+bn+c=0, we get

a=4, b=5 and c=−636

Applying quadratic formula, n = $\frac{(-b\pm \sqrt{b^2}-4ac)}{2}$ and putting values of a, b and c, we get

n= (−5± $\sqrt{25}$−4 (4) $\frac{(-636)}{8}$

$\Rightarrow$ n= (−5 ± $\sqrt{25}$−4 )(4) $\frac{(-636)}{8}$ = (−5± $\sqrt{25}$ +10176)8 = $\frac{-5\pm 101}{8}$

$\Rightarrow$n= $\frac{(-5+101)}{8}$, $\frac{(-5-101)}{8}$

$\Rightarrow$ n=$\frac{96}{8}$, −$\frac{106}{8}$

=12, −$\frac{106}{8}$

We discard negative value of n here because n cannot be in negative, n can only be a positive integer.

Therefore, n=12.

Therefore, 12 terms of the given sequence make sum equal to 636.