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:

S_q = 63q – 3q²

p^th 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, 11^th term is 0.