Question

the sum of four terms of an a.p. is 4 and the sum of product of first and last term and product of middle terms in -38, then find the numbes.

Answer 1

terms are
a, a+d, a+2d, a+3d

4a + 6d = 4
2a + 3d = 2 ---> a = (2-3d)/2

a(a+3d) + (a+d)(a+2d) = -38
a^2 + 3ad + a^2 + 3ad + 2d^2 = 38
2a^2 + 6ad + 2d^2 = -38
2(4 - 12d + 9d^2)/4 + 6d(2-3d)/2 + 2d^2 = -38
(4 - 12d + 9d^2)/2 + 6d(2-3d)/2 + 2d^2 = -38
times 2
4 - 12d + 9d^2 + 12d - 18d^2 + 4d^2 + 76 = 0

solve for d, it comes out as a nice number
d = ± ...
sub back into a = ....