Question

The sum of 3 consecutive terms of AP is 36 and the sum of their squares is 440. Find the terms

Answer 1

Answer:

The 3 consecutive terms of the AP are $12 -2\sqrt{2}$, 12and $12 +2\sqrt{2}$

Explanation:

Let the 3 consecutive terms of the AP be (a-d), (a), and (a+d).

Given that their sum = 36

∴ (a - d) + a + (a + d) = 36

3a = 36

a = 12

Now given that the sum of their squares = 440

∴$(a-d)^{2} +(a)^{2} +(a+d)^{2}$ = 440

$a^{2} +d^{2} -2ad+a^{2} +a^{2} +d^{2}+2ad$ = 440

$3a^{2} +2d^{2}$ = 440

We know a = 12

∴$3(12)^{2} +d^{2}$ = 440

$d^{2}$ = 8

d = $2\sqrt{2}$

∴The 3 consecutive terms are ($12$- $2\sqrt{2}$) , 12 and $12 + 2\sqrt{2}$