The sum of first q terms of an AP is (63q − 3q2). If its pth term is −60, find the value of p.

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

$S_{q}=63 q-3 q^{2}$

$\Rightarrow S_{q-1}=63(q-1)-3(q-1)^{2}$

$=63 q-63-3\left(q^{2}-2 q+1\right)$

$=-3 q^{2}+69 q-66$

Suppose aq denote the qth term of the AP.

$\therefore a_{q}=S_{q}-S_{q-1}$

$=\left(63 q-3 q^{2}\right)-\left(-3 q^{2}+69 q-66\right)$

$=-6 q+66 \quad \ldots(1)$

Now,

$a_{p}=-60$                   (Given)

$\Rightarrow-6 p+66=-60$         [From (1)]

$\Rightarrow-6 p=-60-66=-126$

$\Rightarrow p=21$

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.