Question

Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the prodcut of their means in 5:6.

Answer 1

Let the four parts be (a – 3d), (a – d), (a + d) and (a + 3d).

Then, sum = 56

⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 56

⇒ 4a = 56

⇒ a = 14

It is given that

For $d=2$ Numbers are $8,12,16,20.$

Thus, the four parts are a – 3d, a – d, a + d, a + 3d i.e., 8, 12, 16 and 20.