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Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find its 25th term.

Answer»

Given that first term of AP is a = 10. 

Also given that sum of first 14 terms of AP is 1505.

∴ For n = 14, Sn = 1505.

Let common difference of AP is d.

∴ Sn = 1505, where n = 14.

Sn\(\frac{n}{2}\) [2 + ( − 1)] (By sum formula of first n terms of AP)

⇒ 1505 = \(\frac{14}{2}\) [2 × 10 + (14 − 1)] (∵ Sn = 1505, n = 14 and a = 10)

⇒ 3010 = 14 (20 + 13d)

⇒ 20 + 13d = \(\frac{3010}{14}=215\)

⇒ 13 d = 215 – 20 = 195

⇒ d = \(\frac{195}{13}=15.\)

Hence, common difference of AP is d = 15.

Now, 25th term of AP is a25 = a + (25 – 1)d = 10 + 24 × 15 = 10 + 360 = 370.

(∵ a = 10 & d = 15)

Hence, 25th term of given AP is 370.



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