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Rakesh is much worried about his upcoming assessment on A. P. He was vigorously practicing for the exam but unable to solve some questions. One of these questions is as shown. If the 3rd and the 9th terms of an A.P. are 4 and –8 respectively, then help Rakesh in solving the problem. i) Form an A.P using the given data ii) Find which term of the A.P. is –160? |
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Answer» i) a3 = a + 2d = 4 …… (i) and a9 = a + 8d = –8 ….. (ii) Subtracting (i) from (ii), we get 6d = –12 ⇒ d = –2 d = –2 and a3 = a + 2d = 4 ⇒ a = 4 – 2(–2) = 4 + 4 = 8 So, the AP is 8, 6, 4, 2, 0, -2,…. ii) Here, a = 8 and d = –2 and an = –160 Also, an = a + (n-1)d ⇒ –160 = 8 + (n-1)(–2) ⇒ –160 – 8 = –2n+2 ⇒ n = 85 |
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