Question

The sum of three consecutive terms of an arithmetic progression is 21 and the sum of the squares of these terms is 165 Find the terms​

Answer 1

Let the terms be
a-d, a, a+d

sooo...

a-d+a+a+d=21

3a=21

a=7

and...
$(a - d) {}^{2} + a {}^{2} + (a \ + d) {}^{2}$

$3a {}^{2} + 2d {}^{2} = 165$
$3 \times 49 + 2d { }^{2} = 165$
$2d {}^{2} = 18$
$d {}^{2} = 9 \\ d = \sqrt{9} \\ d = 3$

the numbers are
a=7
a2=10
a3=13