Question

The sum of first q terms of an AP is $(63q-3q^2)$. If its pth term is - 60, find the value of p. Also, find the 11th term of its AP.

Answer 1

Given $S_q=63q-3q^2$

$S_1=63\times 1-3\times 1\times 1=60=a_1$ sum of first 1 term is the first term

$S_2=63\times 2-3\times 2\times 2=114$

$S_2-S_1=114-60=54=a_2$ Sum of first two terms - the sum of the first term is 2 and term.

SO , $d=a_2-a_1=54-60=-6$

Now,

$a_p=-60$

$a+(p-1)d=-60$

$60+(p-1)(-6)=-60$

$p-1=\frac{-60-60}{-6}=20$

$p=21$

Now,

$a_{11}=a+10d=60+10\times -6=60+-60=0$