Question

The sum of first nth terms of an AP is given by Sn =(2nsquare+3n) find the 16th term of the AP

Answer 1

Sn=2n2+3n
an=Sn-Sn-1
=2n2+3n-{2(n-1)2+3(n-1)}
=2n2+3n-{2(n2+1-2n)+3n-3}
=2n2+3n-{2n2+2-4n+3n-3}
=2n2+3n-2n2-2+4n-3n+3
=4n+1
an=4n+1
16 term..
4(16)+1
=64+1
=65..
hence the required answer is 65..

hope this helps!!
cheers!!

Answer 2

sum of first n terms = 2n^2+3n
therefore t1 = S1 = 2*1^2+3*1 = 2+3 = 5
S2 = t1+t2 = 2*2^2+3*2 = 8+6 = 14
therefore t2 = S2-S1 = 14-5 = 9
d = t2-t1 = 9-5 = 4
so, the AP is 5, 9, 13, ....
a=5, d=4, n=16
an = a+(n-1)d
t16 = 5+15*4 = 5+60 = 65