Question

divide 56 into four parts which are the four terms of an AP such that the product of the first and the fourth terms is to the product of the second and the third terms as 5:6

Answer 1

4 no =a-3d,a-d,a+d,a+3d

sum , 4a=56

a=14

(a-3d)(a+3d)/(a-d)(a+d)=5:6

6(a²-9d²)=5(a²-d²)

a²=54d² - 5d²

14² =49d²

d²=14² /7²

d=±14/7=±2

for d=2,

a-3d=14-6=8

a-d=14-2=12

a+d=14+2=16

a+3d=14+6=20

Numbers are 8,12,16,20