the sum of three consecutive terms of an AP is 21,and sum of their square is 165,find these terms
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there terms are:- a-d,a,a+d
a-d+a+a+d=21
3a=21
a=7
(7-d)^2+7+(7+d)^2=165
49+d^2-14d+49+49+d^2+14s=165
147+2d^2=165
2d^2=165-147=18
d^2=18/2=9
d=3,-3
so the terms are
4,7,10 or
10,7,4
a-d+a+a+d=21
3a=21
a=7
(7-d)^2+7+(7+d)^2=165
49+d^2-14d+49+49+d^2+14s=165
147+2d^2=165
2d^2=165-147=18
d^2=18/2=9
d=3,-3
so the terms are
4,7,10 or
10,7,4
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