Question

The product of three consecutive positive integers is $8$ times their sum. The sum of their squares is:

• A. $50$
• B. $77$
• C. $100$
• D. $149$

Answer 1

Answer: (B)

$77$

Let the numbers be $(n-1),n,(n+1)$

According to the given condition,
$(n-1)\left(n\right)(n+1)=8(n+n-1+n+1)$
$\Rightarrow n(n^2-1)=8\left(3n\right)$
$\Rightarrow (n^2-1)=24$
$\Rightarrow n^2=25$
since $n$ is positive integer, $n=5$
Thus required numbers are $4,5,6$
$\therefore 4^2+5^2+6^2=77$