Question

Divide 32 into four parts which are in A.P. such that the product of extremes is to the product of means is 7 : 15.

Answer 1

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

Sum = 32

⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 32

⇒ 4a = 32 ⇒ a = 8

It is given that

`\frac{(a-3d)\,(a+3d)}{(a-d)\,(a+d)}=\frac{7}{15}`

 

`(a^2-9d^2)/(a^2-d^2)=7/15=>(64-9d^2)/(64-d^2)=7/15`

⇒ 128d² = 512

⇒ d² = 4 ⇒ d = ± 2

Thus, the four parts are a – d, a – d, a + d and a + 3d i.e. 2, 6, 10 and 14.