Question

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

Answer 1

Hope this will help you ^_^

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

$\textbf{\underline{Let\;the\;first\;term\;be\;a}}$

$\textbf{\underline{Common\;difference\;be\;d}}$

Here

a = 9

d = 17 - 9 = 8

Sum = 636

${\boxed{\sf\:{Using\;the\;formula}}}$

${\boxed{\sf\:{S_{n}=\dfrac{n}{2}[2a+(n-1)d]}}}$

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

636 × 2 = n{18 + 8n - 8}

1272 = n{10 + 8n}

1272 = 10n + 8n²

8n² + 10n - 1272 = 0

${\boxed{\sf\:{Taking\;2\;as\;common}}}$

2(4n² + 5n - 636) = 0

4n² + 5n - 636 = 0

${\boxed{\sf\:{Splitting\;the\;middle\;term}}}$

4n² + 53n - 48n - 636 = 0

4n(n - 12) + 53(n - 12) = 0

(n - 12) = 0

n = 12

4n + 53 = 0

4n = -53

${\boxed{\sf\:{n=\dfrac{-53}{4}}}}$

$\textbf{\underline{Fractional\;value\;is\;rejected}}$

Hence we get :-

n = 12

$\Large{\boxed{\sf\:{n=12}}}$