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

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One of the basic question from arithmetic progression chapter in which we are to find 3 numbers such that they are in AP if their product is 440 and their sum is 24.

Let a−d,a,a+d are three numbers in AP.
their sum =a−d+a+a+d=24 3a=24
or a=8
Again,
Their product =(a−d)×a×(a+d)=440 a(a^2−d^2)=440 8(8^2−d^2)=440
or d=±3

We have 2 conditions here:
At a=8,d=3
Numbers are: 5,8 and 11
At a=8 and d=−3
Numbers are: 11,8,5.

Let a−d,a,a+d are three numbers in AP.

their sum =a−d+a+a+d=24

3a=24

or a=8

Again,

Their product =(a−d)×a×(a+d)=440

a(a^2−d^2)=440

8(8^2−d^2)=440

or d=±3

We have 2 conditions here:

At a=8,d=3

Numbers are: 5,8 and 11

At a=8 and d=−3

Numbers are: 11,8,5.