The product of four consecutive positive integers is 840 .find the numbers.

I know the answer but don't know how to solve it
the answer is 4,5,6,7
please show me the method.
let four consecutive integers are [a-3d],[a-d],[a+d],[a+3d]
Asked Sep 01, 2014
I may be wrong. its been 30 years since last using algebraic expressions [x(x+1)]*[(x+2)*(x+3)]=840 or [x(x-1)]*[(x-2)*(x-3)]=840 multiplication of Binomial expressions
Answered Sep 02, 2014
The four consecutive numbers are (a-3d),a-d,(a+d),a-3d.
As here difference is 2d.but numbers are consecutive, so difference must be 1.
Therefore, 2d =1 i.e. d=0.5
a^4 - 2.5a^2 - 0.5605=840
Thus gives us a=5.5
Hence, a-3d =5.5 - 1.5 =4
a-d =5.5-0.5 =5
a+d =5.5+0.5 =6
And a+3d =5.5+1.5 =7
Therefore the four numbers are 4,5,6,7
Answered Dec 29, 2017

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