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Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40. |
Answer» Let a be the first term and d be the common difference. Given: 4th term = a4 = 9 Sum of 6th and 13th terms = a6 + a13 = 40 Now, a4 = a + (4-1)d 9 = a + 3d a = 9 – 3d ….(1) And a6 + a13 = 40 a + 5d + a + 12d = 40 2a +17d = 40 2(9 – 3d) + 17d = 40 (using (1)) d = 2 From (1): a = 9 – 6 = 3 Required AP = 3,5,7,9,….. |
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