Question

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

Answer 1

Heya Friend !!

AP ➡️ 9,17,25....
n=?
a=9
Sn=636
d=a2 - a1
d=>17-9
d=>8

We know that sum of n terms
= n/2[2a+(n-1)d]

Now , substituting the values

=> 636 = n/2[2×9+(n-1)8]
=> 636 = n/2[18+8n-8]
=> 636 = n/2[10+8n]
=> 636×2 = 10n+8n^2
=> 1272 = 10n+8n^2
=> 8n^2+10n-1272 = 0
=> 2[4n^2+5n-636] = 0
=> 4n^2+5n-636 = 0
=> 4n^2+53n-48n-636 = 0
=> n(4n+53)-12(4n+53) = 0
=> (n-12) (4n+53) = 0
=> n-12 = 0 , 4n+53 = 0
=> n = 12 , 4n = -53
=> n = 12 , n = -53/4

{ Discarding the value of n as - 53/4 because number of terms cannot be negative or in fraction }

Therefore , n=12

Hence, 12 terms of the AP must be taken to give the sum of 636 .

Answer 2

the given AP is 9 ,17 ,25
here a = 9
and d= 17-9 = 8
so sn = 636
n/ 2 (2× 9 + ( n-1 )× 8 ) = 636
n/ 2 (18+8n -8) = 636
n/2 (10+8 n) = 636
n(5+4n) = 636
4 n^2+ 5 n - 636= 0
4 n^ 2 - 48 n+ 53 n - 636 = 0
4 n ( n- 12)+ 53 ( n-12)= 0
( n-12) (4 n+53)= 0
n-12= 0. or. 4n+53=0
n=12 or n= -53/4
therefore n=12
hence the required no. of term is 12


if it is helpful then please mark it as brainliest☺️