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The Sum of the three terms of an A.P. is 21 and the product of the first, and the third terms exceed the second term by 6, find three terms. |
Answer» Given : The sum of first three terms is 21 To find : The first three terms of AP Assume the first three terms are a - d, a, a + d where a is the first term and d is the common difference So, Sum of first three terms is a - d + a + a + d = 21 3a = 21 a = 7 It is also given that product of first and third term exceeds the second by 6. So, (a - d)(a + d) - a = 6 a2 - d2 - a = 6 Substituting a = 7, 72 - d2 - 7 = 6 d2 = 36 d = 6 or d = - 6 Hence, The terms of AP are a - d, a, a + d which is 1, 7, 13 or 13, 7, 1. |
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