Question

How many terms of the AP:9,17.25,
must be taken to give a sum 636​

Answer 1

Given:-

• An A.P = 9, 17, 25, . . .

• $S_n=636$

To Find:-

• Terms of the AP:9, 17, 25, must be taken to give a sum 636.

Formula used:-

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

Solution:-

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

Here,

• $S_n=636$

• a = 9

• d = 17 - 9 = 8

$\implies\:636=\dfrac{n}{2}[2 \times 9+(n-1)(8)]$

$\implies\:636=\dfrac{n}{2}[18+8n-8]$

$\implies\:636=\dfrac{n}{2}[10+8n]$

$\implies\:636\times 2=n(10+8n)$

$\implies\:8n^2+10n-1272=0$

$\implies\:4n^2+5n-636=0$

By using Middle Term Split,

$\implies\:4n^2+(53-48)n-636=0$

$\implies\:4n^2+53n-48n-636=0$

$\implies\:n(4n+53)-12(4n+53)=0$

$\implies\:(4n+53)(n-12)=0$

$\implies\:4n+53=0 \:and\:n-12=0$

$\implies\:n=\dfrac{-53}{4} \:and\:n=12$

n cannot be negative, then n = 12

Hence, 12 terms of the given A.P must be taken to give a sum of 636.

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