Question

SUM OF THE FOUR CONSECUTIVE NUMBERS IN AN AP IS 32 AND THE RATIO OF THE PRODUCT OF THE FIRST AMD LAST TERMS TO THE PRODUCT OF TWO MIDDLE TERMS IS 7 RATIO 15 FIND THE NUMBER

Answer 1

let the numbers be a, a+d, a+2d and a+3d
a + a+d + a+2d + a+3d = 32
4a + 6d =32 ----> a = (16-3d)/2

a(a+3d)/( (a+d)(a+2d) ) = 7/15
(a^2 +3ad)/(a^2 + 3ad + 2d^2) = 7/15
15a^2 + 45ad = 7a^2 + 21ad + 14d^2
8a^2 + 24ad - 14d^2 = 0

8(16-3d)^2 /4 + 24d(16-3d)/2 - 14d^2 = 0
2(256 - 96d + 9d^2) + 12d(16 - 3d) - 14d^2 = 0
512 - 192d + 18d^2 + 192d - 36d^2 - 14d^2 = 0
-32d^2 = -512

finish it up, you will get 2 different sequences, they both work