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.
RS Aggarwal, Class 10, chapter 5B, question no 8.
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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.