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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. |
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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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