Question

The sum of 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.

Answer:

The pth term of the AP is 21 and the 11th term is 0.

If the sum of 'q' terms is (63q - 3q²),
Sum of 'p' terms = 63p - 3p² and the sum of 'p-1' terms = 63(p-1) - 3(p-1)²
As we know that difference between the sum of 'n' terms and sum of 'n-1' terms gives us the nth term,
(Sum of 'p' terms) - (Sum of 'p-1' terms) = pth term which is given to us to be -60.
Therefore, we have the equation:
63p - 3p² - [63(p-1) - 3(p-1)²] = -60
⇒ 63p - 3p² - 63(p-1) + 3(p-1)² = -60
⇒ 63p - 63(p-1) - 3p² + 3(p² + 1 -2p) = -60
Taking 63 and -3 common, we have:
⇒ 63[p-(p-1)] - 3[p²-(p²+1-2p)] = -60
⇒ 63(p-p+1) - 3(p²-p²-1+2p) = -60
⇒ 63(1) - 3(2p-1) = -60
⇒ 63-6p+3 = -60
⇒ 6p = 126
⇒ p = 21

Similarly, difference between the sum of 11 terms and (11-1) = 10 terms would give us the 11th term.
Therefore, the 11th term:
= 63(11) - 3(11)² - [63(10) - 3(10)²]
= 63(11) - 3(121) - 63(10) +3(100)
= 63(11-10) - 3(121-100)
= 63*1 - 3(21)
= 63 - 63
= 0


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