The sum of three numbers in AP is -3 and their product is 8. Find the numbers
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S3 = -3
3/2(2a+2d) =-3
a+d = -1
and, a × (a+d) ×(a+2d) = 8
a×a^2+2d^2+2ad+ad = 8
a^3 + 2d^2 + 3ad = 8
but,d = -a -1
so,a^3 + 2×{-(a+1)}^2 +3a×-(a+1) = 8
a^3 + 2×(a^2+1+2a) - 3a^2 +3a = 8
a^3 + 2a^2 +4a +2 -3a^2 +3a = 8
a^3 - a^2 + 7a - 8 = 0
solve it.....
3/2(2a+2d) =-3
a+d = -1
and, a × (a+d) ×(a+2d) = 8
a×a^2+2d^2+2ad+ad = 8
a^3 + 2d^2 + 3ad = 8
but,d = -a -1
so,a^3 + 2×{-(a+1)}^2 +3a×-(a+1) = 8
a^3 + 2×(a^2+1+2a) - 3a^2 +3a = 8
a^3 + 2a^2 +4a +2 -3a^2 +3a = 8
a^3 - a^2 + 7a - 8 = 0
solve it.....
Answered by
0
Answer:
the numbers are -4, -1 , 2 or 2,-1,-4
see the attachment for the solution.
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