The sum of 3 consecutive terms of an AP is 6 and their products is -42. Find these terms.
Let the first term be 'a' and the common difference be 'd'.
Then the three consecutive terms will be: a, a+d, a+2d
From the given information:
a + (a+d) + (a+2d) = 6
Simplifying:
3a + 3d = 6
a + d = 2
Also,
a(a+d)(a+2d) = -42
Substituting for a+d:
a(2a+d)(a+2d) = -42
Simplifying:
2a^3 + 6ad^2 + 4a^2d + ad^2 = -42
2a^3 + 7ad^2 + 4a^2d = -42
Dividing both sides by 2 and substituting a+d:
a^3 + 3ad^2 + 2a^2d = -21
Substituting for a+d from the earlier equation:
a^3 + 6a^2d + 4ad^2 = -21
Substituting a+d=2:
a^3 + 12a + 8d^2 = -21
a^3 + 12a + 8(2-a)^2 = -21
Simplifying and solving for 'a':
a^3 - 2a^2 - 29a - 85 = 0
(a-5)(a^2 + 3a + 17) = 0
The quadratic factor has no real roots, so a = 5
Substituting a=5 into the earlier equations:
a+d = 2
5+d = 2
d = -3
Therefore, the three consecutive terms are:
5, 2, -1