Question

The sum of first q terms of an AP is (63q – 3q²⁾. If its p^th term is –60, find the value of p. Also, find the 11^th term of its AP.

Answer 1

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

S_q = 63_q – 3q²

⇒ S_q–1 = 63(q – 1) –3(q – 1)² 

= 63q – 63 – 3 (q² – 2q + 1) 

= –3q² + 69q – 66

Suppose a_q denote the q^th term of the AP

∴ a_q = S_q – S_q–1 

= (63q – 3q²) – (3q² + 69q – 66) 

= –6q + 66       ...(1)

Now, 

a_p = –60     ...(Given)

`\implies` –6p + 66 = –60      ...[From (1)]

`\implies` –6p = –60 – 66 = –126 

`\implies` p = 21 

Thus, the value of p is 21.

Putting q = 11 in (1), we get

a₁₁ = –6 × 11 + 66 = –66 + 66 = 0

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

Answer 2

\[S_q = 63q - 3 q^2 \]

We know

\[ a_q = S_q - S_{q - 1} \]

\[ \therefore a_q = 63q - 3 q^2 - 63\left( q - 1 \right) + 3 \left( q - 1 \right)^2 \]

\[ a_q = 66 - 6q\]

\[\text{ Now}, a_p = - 60\]

\[ \Rightarrow 66 - 6p = - 60\]

\[ \Rightarrow 126 = 6p\]

\[ \Rightarrow p = 21\]

\[ a_{11} = 66 - 6 \times 11 = 0\]