Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one…

Let, the number of balls on one side of the equilateral triangle is $n,$ then the number of balls on each side of the square will be $n-2.$

Then, total balls used in the triangle are $1+2+3+...+n$

Now, the sum of first $n$ natural numbers is $1+2+3+...+n=\frac{n\left(n+1\right)}{2}$

And, the total balls used in the square is $\left(n-2\right)+\left(n-2\right)+\left(n-2\right)+....\left(n-2\right) times={\left(n-2\right)}^2$

From the given condition we can write
$\frac{n\left(n+1\right)}{2}+99={\left(n-2\right)}^2$

$\Rightarrow n\left(n+1\right)+198=2\left(n^2-4n+4\right)$

$\Rightarrow n^2-9n-190=0$

$\Rightarrow \left(n-19\right)\left(n+10\right)=0$

$\Rightarrow n=19$ or $n=-10,$ neglected as $n$ is the number of balls and hence, it can't be negative.

Hence, number of balls in equilateral triangle

$=\frac{n\left(n+1\right)}{2}=\frac{19\cdot 20}{2}=190.$