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The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds these term by 6, find three terms. |
Answer» Let the three terms be a - d, a, a + d a – d + a + a – d = 21 (given) 3a = 21 a = 7 According to question, (a – d) (a + d) = a + 6 a 2 – d2 = a + 6 7 2 – d2 = 7 + 6 49 – d2 = 13 d 2 = 36 d = 6 Therefore, a – d = 7 – 6 = 1 a + d = 7 + 6 = 13 Thus, the three terms are 1, 7 and 13. |
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