Q. 94.7( 12 Votes )

The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165. Find these terms.

Answer :

Let the numbers be (a - d), a, (a + d).

Now, sum of the numbers = 21


(a - d) + a + (a + d) = 21


3a = 21


a = 7


Now, sum of the squares of the terms = 165


(a - d)2 + a2 + (a + d)2 = 165


a2 + d2 - 2ad + a2 + a2 + d2 + 2ad = 165


3a2 + 2d2 + a = 165


Put the value of a = 7, we get,


3(49) + 2d2 = 165


2d2 = 165 - 147


2d2 = 18


d2 = 9


d = 3


If d = 3, then the numbers are 4, 7, 10.


If d = - 3, then the numbers are 10, 7, 4.


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