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

To Find: The three numbers which are in AP.

Given: Sum and sum of the squares of three numbers are 21 and 165 respectively.

Let required number be (a - d), (a), (a + d). Then,

(a - d) + a + (a + d) = 21 3a = 21 a = 7

Thus, the numbers are (7 - d), 7 and (7 + d).

But their sum of the squares of three numbers is 165.

(7 - d)²7²(7 + d)²= 165

49 + d²14d + 49 + d² + 14d = 116

2d² = 18 d² = 9 d = 3

When d=3 numbers are 4, 7, 10

When d= (3) numbers are 10, 7, 4

So,Numbers are 4, 7, 10 or 10, 7, 4.