Question

the sum of three consecutive terms of an ap is 21 and the sum of the squares of these terms is 165 find these terms

Answer 1

let the terms are a-d,a,a+d
sum=a-d+a+a+d=3a
3a=21
a=7
second condition
a2+d2-2ad+a2+a2+d2+2ad=165
3a2+2d2=165
49*3+2d2=165
147+2d2=165
2d2=18
d2=9
d=3
so the series is 4,7,10

Answer 2

let three consecutive terms be a-d, a, a+d
where d is common difference


a-d+a+a-d=21
3a=21

a=7

(a-d)²+a²+(a+d)²=165
a²-2ad+d²+a²+a²+2ad+d²=165
3a²+2d²=165
147+2d²=165
2d²=18
d²=9
d=3

a-d=4
a+d=10

the terms are 4,7,10

hope it is helpful