Question

The sum of first q terms of an AP is (63q − 3q²). If its pth term is −60, find the value of p. Also, find the 11th term of its AP. [CBSE 2013]

Answer 1

Let S_q denote the sum of the first q terms of the AP. Then,

$$\begin{gathered} S_q=63q-3q^2 \\ \Rightarrow S_{q-1}=63\left(q-1\right)-3{\left(q-1\right)}^2 \\ =63q-63-3\left(q^2-2q+1\right) \\ =-3q^2+69q-66 \end{gathered}$$

Suppose a_q denote the q^th term of the AP.

$$\begin{gathered} \therefore a_q=S_q-S_{q-1} \\ =\left(63q-3q^2\right)-\left(-3q^2+69q-66\right) \\ =-6q+66.....\left(1\right) \end{gathered}$$
Now,

$$\begin{gathered} a_p=-60\left(\mathrm{Given}\right) \\ \Rightarrow -6p+66=-60\left[\mathrm{From}\left(1\right)\right] \\ \Rightarrow -6p=-60-66=-126 \\ \Rightarrow p=21 \end{gathered}$$

Thus, the value of p is 21.

Putting q = 11 in (1), we get

$a_{11}=-6\times 11+66=-66+66=0$

Hence, the 11th term of the AP is 0.