The sum of 18 consecutive natural numbers is a perfect square. What is the smallest possible value of this… · ABC26GN1431

Let the 18 consecutive natural numbers be $x,(x+1)(x+$ $2),(x+3), \ldots \ldots,(x+17)$. Then, $x+(x+1)+(x+2)+\ldots \ldots+$ $(x+17)=18 x+(1+2+3+\ldots+17)=18 x+153$. Putting $x=1,2,3,4, \ldots \ldots$. we find that the smallest value of $x$ for which $(18 x+153)$ becomes a perfect square is $x=4$. ∴ Required value $=18 \times 4+153=72+153=225$.