Question

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

Answer 1

Sum of n terms of an ap is given by: S=n/2[2a +(n-1)d]
N=?(need to find out)
S=636
a=9
d=b-a=17-9=8

Substituting in the formula
636=n/2[18+(n-1)8]
1272=n[18+8n-8]
1272=n[10+8n]
1272=10n+8n²

Getting all term to one side:
8n²+10n-1272=0
Diving the whole term by 2

4n²+5n-636=0
Either using splitting the middle term or by the formula of quadratic equation.

We get n=12
Hence the sum of first 12 terms give the total of 636.

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\bf\huge => n = 12$

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

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