Question

Three consecutive positive integers are such that the square of their sum exceeds the sum of their squares by 214. Which are those three integers?

• A. $6,7,8$
• B. $4.5,6$
• C. $7,8,9$
• D. $5,6,7$

Answer 1

The correct option is D $5,6,7$

Let the three consecutive positive integers be $x,x+1\text{and}x+2$.

Given,

$x^2+(x+1{)}^2+(x+2{)}^2=(x+x+1+x+2{)}^2-214$

$\Rightarrow x^2+x^2+1+2x+x^2+4+4x=(3x+3{)}^2-214$

$\Rightarrow 3x^2+6x+5=9x^2+9+18x-214$

$\Rightarrow 3x^2+6x+5=9x^2+18x-205$

$\Rightarrow 6x^2+12x-210=0$

$\Rightarrow x^2+2x-35=0$

$\Rightarrow x^2+7x-5x-35=0$

$\Rightarrow x(x+7)-5(x+7)=0$

$\Rightarrow (x+7)(x-5)=0$

$\therefore x+7=0orx-5=0$

$\Rightarrow x=-7orx=5$

Given that the numbers are positive.

Hence, $x$ cannot be $-7$

$\therefore x=5$

So, $x+1=5+1=6$

and $x+2=5+2=7$

$\therefore$The three consecutive positive integers are $5,6\text{and}7$.