Question

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

Answer 1

a = 9
n = ?
sn = 636
d = 17 - 9 = 8
sn = n/2 ( 2a + (n-1)(d) )
636 = n/2 ( 2(9) + (n-1)(8) )
636 = n/2 ( 18 + 8n - 8 )
636 = n/2 ( 10 + 8n )
1272 = 10n + 8n^2
636 = 5n + 4n^2
4n^2 + 5n - 636 = 0
4n^2 + 53n - 48n - 636 = 0
n(4n + 53) - 12(4n + 53) = 0
(n-12)(4n+53) = 0
4n = -53
n = -53/4
n = 12
Terms cannot be negative so n = 12.
12 terms must be taken for the sum to be 636.

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\bf\huge => n = 12$

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

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