JEE Main 2019 — Mathematics Algebra

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

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

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

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

From the given condition we can write

$$\frac{n(n+1)}{2}+99={(n-2)}^{2}$$2n(n+1)​+99=(n−2)2

$$\Rightarrow n(n+1)+198=2({n}^{2}-4n+4)$$⇒n(n+1)+198=2(n2−4n+4)

$$\Rightarrow {n}^{2}-9n-190=0$$⇒n2−9n−190=0

$$\Rightarrow (n-19)(n+10)=0$$⇒(n−19)(n+10)=0

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

Hence, number of balls in equilateral triangle

$$=\frac{n(n+1)}{2}=\frac{19\cdot 20}{2}=190.$$=2n(n+1)​=219⋅20​=190.