find the three consecutive terms in an a.p whose sum is 18 and the sum their squares is 140
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Hello !
n + ( n + 1 ) + ( n + 2 ) = 18
3n + 3 = 18
3n = 18 - 3 = 15
n = 15 ÷ 3 = 5
5 ; 6 & 7
5² + 6² + 7²
25 + 36 + 49
110
n + ( n + 1 ) + ( n + 2 ) = 18
3n + 3 = 18
3n = 18 - 3 = 15
n = 15 ÷ 3 = 5
5 ; 6 & 7
5² + 6² + 7²
25 + 36 + 49
110
velssraju:
wrong
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