# the sum of three numbers in an arithmetic progression is 15 and their product is 105. Find the numbersIt has to do with Arithmetic Progression and the formula

### 1 Answer | Add Yours

Let x,x+d and x+2d be the three numbers.

Then x+x+d+x+2d=15 or 3x+3d=15 ==>x+d=5

We are also given x(x+d)(x+2d)=105

Substituting `5-x` for d we get:

x(x+5-x)(x+2(5-x))=105

x(5)(-x+5)=105

`-5x^2+50x=105`

`5x^2-50x+105=0`

`5(x-3)(x-7)=0` so x=3 or x=7

If x=7 then d=-2 and we have 7,5,3 as the numbers.

If x=3 then we have 3,5,7 as the numbers.

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**The sum of 3,5,7 is 15 and the product is 105 so the required numbers are 3,5,7.**

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