Q. 75.0( 2 Votes )

# The sum of the 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 this AP.

Answer :

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

Given, Sum of q terms:

S_{q} = 63q – 3q^{2}

p^{th} term = Sum of p terms – Sum of (p – 1) terms

⇒ –60 = 63p – 3p^{2} – (63(p – 1) – 3(p – 1)^{2})

⇒ –60 = 63p – 3p^{2} – (63p – 63 – 3(p^{2} – 2p + 1))

⇒ –60 = 63p – 3p^{2} – (63p – 63 – 3p^{2} + 6p – 3)

⇒ –60 = 63p – 3p^{2} – 63p + 63 + 3p^{2} – 6p + 3

⇒ –60 = –6p + 66

⇒ 6p = 126

⇒ p = 31

Also,

a_{11} = S_{11} – S_{10}

⇒ a_{11} = 63(11) – 3(11)^{2} – (63(10) – 3(10)^{2})

⇒ a_{11} = 693 – 363 – 630 + 300

⇒ a_{11} = 0

Hence, 11^{th} term is 0.

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