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=Â First term
d=Â Common difference
n=Â 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.