Question

Divide 56 in four parts in A.P such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is 5 : 6​

Answer 1

➢ Answer :

Four parts in A.P are 8, 12, 16, 20 or 20, 16, 12, 8.

➢ Given :

Sum of four parts in A.P = 56

Ratio of the product of extremes to the product of means of the 4 parts = 5 : 6.

➢ To Find :

Four parts in A.P.

➢ How To Find :

Let the four parts in AP be a + 3d, a + d, a - d, a - 3d

We are given that their sum = 56

⇝ a + 3d + a + d + a - d + a - 3d = 56

⇝ 4a = 56

⇝ a = 14

Now, we are also given that the ration of the product of their extremes to the product of means is 5 : 6,

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

Substituting the value of a, we get,

⇝ (14 - 3d)(14 + 3d)/(14 - d)(14 + d) = 5/6

⇝ 14² - (3d)²/196 - d² = 5/6

⇝ 196 - 9d²/196 - d² = 5/6

⇝ 1176 - 54d² = 980 - 5d²

⇝ - 49d² = - 196

⇝ d² = 4

⇝ d = √4

⇝ d = ±2

___________________________

∴ When d is 2, four parts in A.P are :

8, 12, 16, 20

When d is -2, four parts in A.P are :

20, 16, 12, 8