Question

Choose the correct answer in each of the following questions:

The sum of first n terms of an AP is (3n² + 6n). The common difference of the AP is [CBSE 2014]

(a) 6 (b) 9 (c) 15 (d) −3

Answer 1

Let S_n denotes the sum of first n terms of the AP.

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

Let d be the common difference of the AP.
$$\begin{gathered} \therefore d=a_n-a_{n-1} \\ =\left(6n+3\right)-\left[6\left(n-1\right)+3\right] \\ =6n+3-6\left(n-1\right)-3 \\ =6 \end{gathered}$$

Thus, the common difference of the AP is 6.

Hence, the correct answer is option A.