Three consecutive positive integers are such that the square of their sum exceeds the sum of their squares by 214. The average of the numbers is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6
The square of a sum is the sum of the squares plus twice the sum of the products in pairs: \((p+q+r)^2=p^2+q^2+r^2+2(pq+qr+rp)\). So the excess of 214 is twice the sum of the pairwise products. Taking the middle number as \(n\), the numbers are \(n-1\), \(n\), \(n+1\) and their average is \(n\).
The excess is twice the pairwise products: \(2(pq+qr+rp)=214\), so \(pq+qr+rp=107\)
The pairwise products of the three numbers add up to \((n-1)n+n(n+1)+(n-1)(n+1)=3n^2-1\)
Setting this equal to 107: \(3n^2=108\), so \(n^2=36\) and \(n=6\)
The numbers are 5, 6 and 7, and their average is 6.