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.

LetÂ aâˆ’d,a,a+dÂ are three numbers in AP.
their sumÂ =aâˆ’d+a+a+d=24 3a=24
orÂ a=8
Again,
Their productÂ =(aâˆ’d)Ã—aÃ—(a+d)=440 a(a^2âˆ’d^2)=440 8(8^2âˆ’d^2)=440
orÂ d=Â±3

LetÂ aâˆ’d,a,a+dÂ are three numbers in AP.

their sumÂ =aâˆ’d+a+a+d=24

3a=24

orÂ a=8

Again,

Their productÂ =(aâˆ’d)Ã—aÃ—(a+d)=440

a(a^2âˆ’d^2)=440

8(8^2âˆ’d^2)=440

orÂ d=Â±3

We haveÂ 2Â conditions here:

AtÂ a=8,d=3

Numbers are:Â 5,8Â andÂ 11

AtÂ a=8Â andÂ d=âˆ’3

Numbers are:Â 11,8,5.