Question

divide 24 in three parts such that they are in AP and their product is 440.

Answer 1

First part = x
Second part = x+d
Third part = x+2d
The two equations are
x + (x+d) + (x+2d) = 24
and x(x+d)(x+2d) = 440
Simplifying the first equation:
x + (x+d) + (x+2d) = 24
x + x + d + x + 2d = 24
3x + 3d = 24
x + d = 8
d = 8-x
Substituting in the second equation:
x(x+d)(x+2d) = 440
x(x+8-x)[x+2(8-x)] = 440
x(8)[x+16-2x] = 440
8x(16-x) = 440
x(16-x) = 55
16x-x² = 55
-x²+16x-55 = 0
x²-16x+55 = 0
(x-11)(x-5) = 0
x-11 = 0 ; x-5 = 0
x = 11; x = 5
d = 8-x
d = 8-x
d = 8-11
d = 8-5. ; d = 3
One solution: First part = x = 11
Second part = x+d = 11+(-3) = 8
Third part = x+2d = 11+2(-3) = 11-6 = 5
Thus 11, 8, and 5

Answer 2

let the no.s be a-d , a , a+ d
Sum of these is 24 ,so a-d + a +a+d= 3a
then 3a=24,a=8
product of them is 440 ,a(a+d)(a-d)=440
a(a2-d2)=440
putting value of a=8
8(64 - d2) = 440
64 - d2 =55
d2 = 9
d=3
so no.s are 8 -3, 8 , 8 +3
=5, 8, 11