Question

how many terms of the ap 9 17 25 must be taken so that their sum is 636?

Answer 1

You Ans Should Be...

s=636
a=9
d=17-9=8
n=?

s=n/2 {2a+(n-1)d}

636=n/2{2×9+(n-1)8}
636×2=n(18+8n-8)
1272=18n+8n^2-8n
1272=8n^2+10n
8n^2+10n-1272=0

D=b^2-4.a.c

= 10^2-4.8.-1272
= 100+40704
= 40804

x= -b+-√D / 2a

= -10+-√40804 / 2×8

= -10+202 / 16 , -10-202 / 16

= 192 /16 , -212 / 16

= 12 , -13.25

So Ans Should Be 12

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\bf\huge => n = 12$

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

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