Question

The sum of the first three consecutive terms of an A.P. is 9 and the sum of their squares is 35.
Find Sn​

Answer 1

solution

Let the three numbers in the AP be a-d, a and a+d.

So a-d+a+a+d = 3a = 9 or a = 3.

Now (a-d)^2+a^2+(a+d)^2 = 35, or

a^2–2ad+d^2+a^2+a^2+2ad+d^2 = 35, or

3a^2+2d^2 = 35, or

3x3^2+2d^2 = 35, or

2d^2 = 35–27 = 8, or d = 2.

So the terms are: 1, 3 and 5. The sum is already in the question as 9, Which other sum are you referring to?