3 consecutive numbers of a series are in A.P with d=2, smallest number?
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the smallest number is 1
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3 consecutive numbers: n, n+1, n+2
Their sum is 3n+3.
Their product is n(n+1)(n+2) = n^3 + 3n^2 + 2n
sum = product, therefore 3n+3 = n^3 + 3n^2 + 2n
Now rearrange that equation to n^3 + 3n^2 - n - 3 = 0
Factorise the cubic equation in order to solve it: (n + 3) (n + 1) (n - 1) = 0
Therefore, the three solutions are n= -3; n=-1; n=1
Their sum is 3n+3.
Their product is n(n+1)(n+2) = n^3 + 3n^2 + 2n
sum = product, therefore 3n+3 = n^3 + 3n^2 + 2n
Now rearrange that equation to n^3 + 3n^2 - n - 3 = 0
Factorise the cubic equation in order to solve it: (n + 3) (n + 1) (n - 1) = 0
Therefore, the three solutions are n= -3; n=-1; n=1
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