From the question, we deduce that 18 is the third integer. Let 18 - 2a ; 18 - a ; 18 + a ; 18 + 2a be the remaining integers. We have :

\(\dfrac{\left(18-2a\right)^2+\left(18-a\right)^2+18^2+\left(18+a\right)^2+\left(18+2a\right)^2}{5}=374\)

\(\Leftrightarrow\dfrac{18^2-72a+4a^2+18^2-36a+a^2+18^2+18^2+36a+a^2+18^2+72a+4a^2}{5}=374\)

\(\Leftrightarrow\dfrac{18^2.5+10a^2}{5}=374\Leftrightarrow18^2+2a^2=374\Leftrightarrow2a^2=50\Leftrightarrow a^2=25\Leftrightarrow a=\pm5\)

The answer is : \(18+5\times2=28\)

From the question, we deduce that 18 is the third integer. Let 18 - 2a ; 18 - a ; 18 + a ; 18 + 2a be the remaining integers. We have :

(18−2a)2+(18−a)2+182+(18+a)2+(18+2a)25=374(18−2a)2+(18−a)2+182+(18+a)2+(18+2a)25=374

⇔182−72a+4a2+182−36a+a2+182+182+36a+a2+182+72a+4a25=374⇔182−72a+4a2+182−36a+a2+182+182+36a+a2+182+72a+4a25=374

⇔182.5+10a25=374⇔182+2a2=374⇔2a2=50⇔a2=25⇔a=±5⇔182.5+10a25=374⇔182+2a2=374⇔2a2=50⇔a2=25⇔a=±5

The answer is : 18+5×2=28