Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are added to t

Question

Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls used in forming the equilateral triangle, then all these balls can be arranged in a square, whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number of balls used to form the equilateral triangle is

TestHub · Accepted Answer

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 $1 + 2 + 3 + . . . + n = \frac{n (n + 1)}{2}$

And, the total balls used in the square is $(n - 2) + (n - 2) + (n - 2) + . . . . (n - 2) t i m e s = {(n - 2)}^{2}$

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

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

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

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

$\Rightarrow n = 19$ 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

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

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