# 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.

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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