We have given a question from arithmetic progression chapter in which we have been asked to find the nth term and the 15th term of the AP, if the sum of the first n terms of an AP is (3n^2+6n).

Book – RS Aggarwal, Class 10, chapter 5C, question no 4

We know, nth term Tn=a+(n−1)d

Where

a= First term

d= Common difference

n= number of terms

Tn=nth term

Given: Sn=3n2+6n

S1=3(1)2+6×1=3+6=9

S2=3(2)2+6×2=12+12=24

T2=S2−S1=24−9=15

d=15−9=6

a=9

Tn=a+(n−1)d=9+(n−1)×6

=9+6n−6=3+6n

=6n+3

T15=6×15+3

=90+3

=93.