SOLUTION: Find four consecutive integers such that the square of the 3rd number is 154 more than the product of the first two numbers

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Question 18398: Find four consecutive integers such that the square of the 3rd number is 154 more than the product of the first two numbers
Answer by venugopalramana(3286) About Me  (Show Source):
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Find four consecutive integers such that the square of the 3rd number is 154 more than the product of the first two numbers
let the 1st.number =x
2nd.no.=x+1
3rd.no.=x+2
4th.no.=x+3
square of 3rd.no.=(x+2)^2=x^2+4x+4
product of 1st.&2nd.nos=x(x+1)
154 more than above=x(x+1)+154=x^2+x+154
hence x^2+4x+4=x^2+x+154
x^2+4x-x^2-x=154-4=150
3x=150
x=150/3=50
hence the 4 numbers are 50,51,52,53