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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)   (Show Source):
You can put this solution on YOUR website! 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
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