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.

Answer 1

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,

S_n = ${\displaystyle \frac{n}{2}[2a+(n-1)d]}$

∴ S_n = ${\displaystyle \frac{n}{2}[2\times 8+(n-1)20]}$

⇒ S_n = ${\displaystyle \frac{n}{2}(16+20n-20)}$

⇒ S_n = ${\displaystyle \frac{n}{2}(20n-4)}$

∴ S_n = 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,

S_2n = ${\displaystyle \frac{2n}{2}[2a+(2n-1)d]}$

⇒ S_2n = n[2(– 30) + (2n – 1)(8)]

⇒ S_2n = n[– 60 + 16n – 8)]

⇒ S_2n = 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

⇒ S_n = S_2n    ...[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.