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Question

(A) Let the number of rows in the equilateral triangle be $n$. The total number of balls in the triangle is given by the sum of the first $n$ natural

Vedclass · Accepted Answer

$190$

Vedclass · Answer

(A) Let the number of rows in the equilateral triangle be $n$. The total number of balls in the triangle is given by the sum of the first $n$ natural numbers: $S = \frac{n(n+1)}{2}$.According to the problem,if $99$ more balls are added,the total number of balls becomes $S + 99 = \frac{n(n+1)}{2} + 99$.These balls form a square with side length $(n-2)$. Thus,the total number of balls is $(n-2)^2$.Equating the two expressions: $\frac{n(n+1)}{2} + 99 = (n-2)^2$.Multiplying by $2$: $n^2 + n + 198 = 2(n^2 - 4n + 4)$.$n^2 + n + 198 = 2n^2 - 8n + 8$.Rearranging the terms: $n^2 - 9n - 190 = 0$.Factoring the quadratic equation: $(n - 19)(n + 10) = 0$.Since $n$ must be positive,$n = 19$.The number of balls used to form the triangle is $\frac{19(19+1)}{2} = \frac{19 	imes 20}{2} = 190$.

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