# The sum of the 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 this AP.

Let the first term be ‘a’ and common difference be ‘d’.

Given, Sum of q terms:

Sq = 63q – 3q2

pth term = Sum of p terms – Sum of (p – 1) terms

–60 = 63p – 3p2 – (63(p – 1) – 3(p – 1)2)

–60 = 63p – 3p2 – (63p – 63 – 3(p2 – 2p + 1))

–60 = 63p – 3p2 – (63p – 63 – 3p2 + 6p – 3)

–60 = 63p – 3p2 – 63p + 63 + 3p2 – 6p + 3

–60 = –6p + 66

6p = 126

p = 31

Also,

a11 = S11 – S10

a11 = 63(11) – 3(11)2 – (63(10) – 3(10)2)

a11 = 693 – 363 – 630 + 300

a11 = 0

Hence, 11th term is 0.

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