the sum of these consecutive terms of an ap is 21 and the sum of the squares of these terms is 165 find these terms
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hello,
we can assume the AP to be (a-d),a,(a+d),with common difference of d
sum of 3 terms is 21
a-d+a+a+d=21
3a=21
a=7
now,sum squares is 165
(a-d)²+a²+(a+d)²=165
a²+d²-2ad+a²+a²+d²+2ad=165
3a²+2d²=165
3×7²+2d²=165
147+2d²=165
2d²=18
d²=9
d=-3 or +3
now,
the three term can be
4,7,10 with common difference 3
hope this helps,if u like it please mark it as brainliest
we can assume the AP to be (a-d),a,(a+d),with common difference of d
sum of 3 terms is 21
a-d+a+a+d=21
3a=21
a=7
now,sum squares is 165
(a-d)²+a²+(a+d)²=165
a²+d²-2ad+a²+a²+d²+2ad=165
3a²+2d²=165
3×7²+2d²=165
147+2d²=165
2d²=18
d²=9
d=-3 or +3
now,
the three term can be
4,7,10 with common difference 3
hope this helps,if u like it please mark it as brainliest
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