Q:
Three consecutive integers form the lengths of a right-angled triangle. How many sets of such three consecutive integers is/are possible?
Answer & Explanation
Answer: A) Only one
Explanation: let n-1, n, n+1 be 3 consecutive integers
So
(n+1)^2= n^2+ (n-1)^2
(n+1)^2-(n-1)^2= n^2
4n = n^2
So n = 0 or n = 4
n can’t be 0 as n-1 will be negative then
So 3,4 and 5 is the only triplet formed.
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