Question

If the sum of four numbers in AP be 96 and that of the product of the means to the product of the extremes is 35:27, then find the numbers.

Answer 1

Take the four numbers as (a-3d),(a-d),(a+d),(a+3d) in AP with common difference as 2d.

Now given that Sum of four numbers = 96

i.e (a-3d)+(a-d)+(a+d)+(a+3d) = 96

=> 4a = 96

=> a = 24 ----------(1)

Now to find d use the other condition.

i.e Product of means/Product of Extremes = 35/27

=> (a-d)(a+d) / (a-3d)(a+3d) = 35/27

=> a² - d² / a² -9d² = 35/27 ---------------(2)

Now find the value of d using equation (1) and (2)

You will get d = 4

So the four numbers are:

a-3d = 24-3*4 = 12

a-d = 24 - 4 = 20

a+d = 24+4 = 28

a+3d = 24+12 = 36

Hence the four numbers are 12,20,28,36