Question

The sum of three consecutive terms of an A. P. is 21 and the sum of their squares is 165. Find these terms.

Answer 1

Let the terms are a - d, a, a + d.
Sum = a - d + a + a + d = 3 a
3 a = 21
a = 7
Second condition,
$a^2$ + $d^2$ - 2ad + $a^2$ + $a^2$ + $d^2$ + 2ad = 165
3$a^2$ + 2$d^2$ = 165
49$\times$3 + 2$d^2$ =165
147 + 2$d^2$ = 165
2$d^2$ = 18
$d^2$ = 9
d = 3
So the series is 4,7,10