Arithmetic progression.?

Question

three terms are in Arithmetic progression. if their sum is 21 and the sum of their squares is 149, what are the numbers?

Answer 1

They are 6, 7, and 8.

To see why, recall that an arithmetic progression {a_0, a_1, ..., a_n, ...} is in the form a_n =m*n + b for some constants m and b. Our first three terms are a_0 = b, a_1 = m+b, and a_2 = 2m+b.

The sum is a_0 + a_1 + a_2 = 3*(m+b), and the sum of the squares is a_0^2 + a_1^2 + a_2^2 = 5*m^2 + 6*m*b + 3*b^2, which can be rewritten as 2*m^2 + 3*(m+b)^2. We are given that the sum is 21 and the sum of the squares is 149, so m+b = 7 and m^2 = 1. This means (m,b) = (-1,8) or (1,6), so that (a_0, a_1, a_2) = (6,7,8) or (8,7,6).

Answer 2

Terms in AP differ by a common difference d, so say the numbers are a - d, a, a + d, Their sum is 3a = 21, so a = 7.

Our numbers are 7 - d, 7 , 7 + d. The sum of their squares is (49 - 14d + d^2) + 49 + (49 + 14d + d^2) = 149, and this simplifies to 2d^2 = 2, so d = +/-1.

Both d = -1 and d = 1 tell us that the numbers are 6, 7, and 8.