InterviewSolution
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Find the 12th term from the end of the following arithmetic progressions:(i) 3, 5, 7, 9, ..... 201(ii) 3, 8, 13, ....., 253(iii) 1, 4, 7, 10, ......., 88 |
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Answer» (i) 3, 5, 7, 9, ..... 201 a = 3, d = 5 - 3 = 2, an = 201 an = a +(n -1)d 201 = 3 +2n – 2 N = 100 Now, we have to find 12th term from the last that means, 100th – 11 = 89th term Then, a89 = a + (89 – 1)d = 3 + 88 x 2 = 179 Hence, the 12th term from the end of the A.P. is 179. (ii) 3, 8, 13, ....., 253 a = 3, d = 8 - 3 = 5 an = 253 a + (n - 1)d = 253 3 + (n - 1)5 = 253 n = 51 Now, we have to find 12th term from the last that means, 51th -11 = 40th term Then, a40 = a + (n -1)d = 3 + 39(5) = 198 Hence, the 12th term from the end of the A.P. is 198 (iii) 1, 4, 7, 10, ......., 88 a = 1, d = 4 -1 = 3 an = 88 a + (n -1)d = 88 1 + (n -1)3 = 88 n = 30 Now, we have to find 12th term from the last that means, 30th - 11 = 19th term a19 = a + 18d = 1 + 18(3) = 55 Hence, the 12th term from the end of the A.P. is 198. |
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