Question

How many terms of the AP : 9,17,25,....must be taken to give asum of 636

Answer 1

a=9, d=17-9=8, Sn=636
Sn=n/2 [(2a+(n-1) d]
636= n/2 [ 2*9+(n-1)*8]
636= n/2 [18+8n-8]
636*2= n (10+8n)
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, n=-53/4
n=-53/4 is not possible
So, n=12

Answer 2

here
first term a=9
common difference=8
we know
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)
636×2=n(10+8n)
1272=10n+8n^2
or8n^2+10n-1272=0
2(4n^2+5n-636)=0
4n^2+5n-636=0/2
since the equation is quadratic ,therefore it will have two roots
Now,
here a=4,b=5,c=636
n=-b+-√b^2-4ac/2a
n=-5+-√5^2-4×4×-636/2×4
n=-5+-√25+10176/8
n=-5+-√10201/8
n=-5+-101/8
Either
n=-5+101/8. or. n=-5-101/8
n=96/8. or. n=-106/8
n=12. or. n=-13.25
Since number of terms cannot be negative ,Therefore n=-13.25 is rejected
Thus ,number of terms required to give the sum =12