Question

How many terms of the AP: 9,17,25,... must be taken to get a sum of 636

Answer 1

Hey !!
A.P. -
9, 17, 25, .....

a=9 ; d=a₂-a₁=17-9=8
Sₙ = n/2[2a+(n-1)d]
636 = n/2[2*9+(n-1)8]
636 = n*[9+(n-1)4]
636 = n*[9+4n-4]
636 = 4n²+5n

4n²+5n-636=0
n= [-5±√5²-4(4)(-636)]/2*4
n= [(-5±√10,201)/8]
n= [(-5±101)/8]
n= (-5+101)/8
n= 96/8
n= 12

{Neglecting negative value, as no. of terms can be negative}

Hence, number of terms = 12 Ans.

Hope it helps :)

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\bf\huge => n = 12$

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

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