Question

How many terms of the AP : 45, 39, 33, must be taken so that their sum is 180?

Answer 1

The given sequence of AP is 45, 39, 33, …………..

Let AP has n terms

Here, a – 45, d = 39 – 45 = – 6 and S_n = 180 (given)

S_n = 180

⇒ \(\frac{n}{2}\) [2a + (n – 1)d] = 180

⇒ \(\frac{n}{2}\)[2 × 45 + (n – 1) × (-6)] = 180

⇒ \(\frac{n}{2}\) [90 – 6n + 6] = 180

⇒ \(\frac{n}{2}\)[96 – 6n] = 180

⇒ 96n – 6n² = 360

⇒ 6n² – 96n + 360 = 0

⇒ n² – 16n + 60 = 0

⇒ n² – (10 + 6)n + 60 = 0

⇒ n – 10n – 6n + 60 =0

⇒ n(n – 10) – 6(n – 10) = 0

⇒ (n – 10) (n – 6) =0

⇒ n = 10 or n = 6

Hence, 10 or 6 terms must be taken.