Question

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

Answer 1

Sum=636.
a=9.
d=17-9=8.
Therefore sum = n/2[2a + (n - 1)d].
636= n/2[2*9+(n-1)8].
636= n/2[18+(n-1)8].
636*2= 18n+8n^2 -8n.
636^2= 8n^2 +10n. (Dividing both side by 2)
636 = 4n^2 +5n.
4n^2 +5n -636=0.
4n^2 -48n +53n -636=0.
4n(n-12) + 53( n+12) =0.
(4n+53) (n-12)=0.( We'll discard 4n+53 as it will be negative)
n-12 =0.
So n=12.
12 terms of the AP 9,17,25....will be taken to give a sum of 636.

Hope this helps.

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\bf\huge => n = 12$

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

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