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
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