Question

how many terms of the AP:9, 17, 25,... must be taken to give a sum of 636​

Answer 1

$\bf{\underline{Question:-}}$

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

$\bf{\underline{Given:-}}$

• First term (a) = 9
• Common difference (d) = a2 - a1 = 17 - 9= 8
• Sum (Sn)= 636

$\bf{\underline{Formula:-}}$

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

$\bf{\underline{Solution:-}}$

Substituting values in formula

$\bf → S_n = \frac{n}{2}[2a+(n-1)d]$

$\bf →636 = \frac{n}{2}[2(9)+(n-1)8$

$\bf →636=\frac{n}{2}[18 + 8n - 8]$

$\bf →636 = \frac{n}{2}[8n + 10]$

$\bf →4n^2 + 5n - 636 = 0$

$\bf →4n^2 + (53-48)n - 636=0$

$\bf → 4n^2 + 53n - 48n - 636 = 0$

$\bf →4n ( n - 12) + 53(n-12)=0$

$\bf →(4n + 53 ) ( n - 12 ) = 0$

$\bf →4n + 53 = 0$

$\bf →4n = -53$

$\bf → n = \frac{-53}{4}$

Either

$\bf →n - 12 = 0$

$\bf →n= 12$

$\bf{\underline{Therefore:-}}$

• - 53/4 is negative So, we neglect it

$\bf{\underline{Hence:-}}$

• 12 term are needed to give the sum of an A.P 636

Answer 2

Answer:

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