Question

how many terms of an A.P. 9 , 17 , 25.... must be taken to give a sum of 636?

Answer 1

Sn =636
a=9
d=8
Sn = n/2[2a+(n-1)d]
Sn=n/2[18+8n-8]
=n/2[10+8n]
=5n+4n²=636

4n²+5n-636=0
by quadratic formula
n=12


hope it helps

Answer 2

$\bf\huge\boxed{\boxed{\bf\huge\:Hello\:Mate}}}$

$\bf\huge Let: first\: term\; be\: a \:and\: CD\: = 17 - 9 = 8$

$\bf\huge => S_{n} = 636$

$\bf\huge => \frac{N}{2}[2a + (n - 1)d] = 636$

$\bf\huge => \frac{N}{2}[2\times 9 + (n - 1)8] = 636$

$\bf\huge => \frac{N}{2} (8n - 10) = 636$

$\bf\huge => n(4n + 5) = 636$

$\bf\huge => 4n^2 + 5n + 636 = 0$

$\bf\huge => n = \frac{-5 + \sqrt{25 - 4\times 4\times -636}}{2\times 4}$

$\bf\huge = \frac{-5 + \sqrt{25 + 10176}}{8}$

$\bf\huge = \frac{- 5 + \sqrt{10201}}{8}$

$\bf\huge = \frac{-5 + 101}{8}$

$\bf\huge = \frac{96}{8} , \frac{-106}{8}$

$\bf\huge = 12 , \frac{-53}{4}$

$\bf\huge But\: n \:cannot\: be\: Negative$

$\bf\huge => n = 12$

$\bf\huge Hence\:Sum\: of\: 12\: terms\: is\: 636$

$\bf\huge\boxed{\boxed{\:Regards=\:Yash\:Raj}}}$