This is the basic and conceptual question from Chapter name- Arithmetic Progression

Chapter number- 9

Exercise 9.6

In this question we have been given that The sum of the first n terms of an A.P. is 3n2 + 6n.

Now we have to Find the nth term of this A.P.

CBSE DHANPAT RAI publications

Class:- 10th

Solutions of CBSE Mathematics

Question 38

Given S

_{n}= 3n^{2}+ 6nOn putting n = 1, we get the first term(a), S

_{1}= a = 3(1)^{2 }+ 6(1) = 9On putting n = 2 gives S

_{2}= a + a + d = 3(2)^{2 }+ 6(2) = 24=> d = 24 – 2a

=> d = 24 – 18 = 6

By using the formula of nth term of an A.P.

a

_{n}= a + (n – 1)dSo,

= 9 + (n – 1)6

= 9 + 6n – 6

= 6n + 3

Hence, the nth term of the given A.P. is 6n + 3.