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Question
The sum of first n terms of an A.P. whose first term is 8 and the common difference is 20 equal to the sum of first 2n terms of another A.P. whose first term is – 30 and the common difference is 8. Find n.
Solution
Given that, first term of the first AP(a) = 8
And common difference of the first AP(d) = 20
Let the number of terms in first AP be n.
∵ Sum of first n terms of an AP,
Sn = `n/2[2a + (n - 1)d]`
∴ Sn = `n/2[2 xx 8 + (n - 1)20]`
⇒ Sn = `n/2 (16 + 20n - 20)`
⇒ Sn = `n/2(20n - 4)`
∴ Sn = n(10n – 2) ...(i)
Now, first term of the second AP(a’) = – 30
And common difference of the second AP(d’) = 8
∴ Sum of first 2n terms of second AP,
S2n = `(2n)/2[2a + (2n - 1)d]`
⇒ S2n = n[2(– 30) + (2n – 1)(8)]
⇒ S2n = n[– 60 + 16n – 8)]
⇒ S2n = n[16n – 68] ...(ii)
Now, by given condition,
Sum of first n terms of the first AP = Sum of first 2n terms of the second AP
⇒ Sn = S2n ...[From equations (i) and (ii)]
⇒ n(10n – 2) = n(16n – 68)
⇒ n[(16n – 68) – (10n – 2)] = 0
⇒ n(16n – 68 – 10n + 2) = 0
⇒ n(6n – 66) = 0
⇒ n = 11 ...[∵ n ≠ 0]
Hence, the required value of n is 11.
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