Question

The sum of the first n terms of an AP is (3n² + 6n). Find the nth term and the 15th term of this AP. [CBSE 2014]

Answer 1

Let S_n denotes the sum of first n terms of the AP.
$$\begin{gathered} \therefore S_n=3n^2+6n \\ \Rightarrow S_{n-1}=3{\left(n-1\right)}^2+6\left(n-1\right) \\ =3\left(n^2-2n+1\right)+6\left(n-1\right) \\ =3n^2-3 \end{gathered}$$

∴ n^th term of the AP, a_n
$$\begin{gathered} =S_n-S_{n-1} \\ =\left(3n^2+6n\right)-\left(3n^2-3\right) \\ =6n+3 \end{gathered}$$

Putting n = 15, we get

$a_{15}=6\times 15+3=90+3=93$

Hence, the n^th term is (6n + 3) and 15th term is 93.