How many billiard balls you need to arrange them into equilateral triangle with 25 rows in it?

25 The number of balls to make an equilateral triangle with the sides of length n is expressed by the formula n(n+1)/2 So to express the number of balls you have is b = n(n+1)/2 + 4 With the larger arrangement you have b = (n+1)(n+2)/2 - 3 Both of those qualities are equal to each other, so set an equation where they're equal. Then solve for n n(n+1)/2 + 4 = (n+1)(n+2)/2 - 3 Distribute the n term on the left (n^2 + n)/2 + 4 = (n+1)(n+2)/2 - 3 Distribute the /2 on the left. 0.5n^2 + 0.5n + 4 = (n+1)(n+2)/2 - 3 Multiply (n+1)(n+2) on right, then distribute the /2 0.5n^2 + 0.5n + 4 = 0.5n^2 + 1.5n + 1 - 3 Subtract 0.5n^2 + 0.5n from both sides 4 = n + 1 - 3 Add 2 to each side 6 = n So the original triangle had sides with a length of 6, for a total number of balls of 6(6+1)/2 = 21 with 4 extra giving 25 balls. Let's check with the next larger triangle 7(7+1)/2 = 28 with 3 balls shortage. Which 25 balls would make happen. So the number of balls in the set is 25

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325

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total number of billiard balls

sum of AP=1 ,2,3,4,25

sum= 25×26/2= 325

hence at least 325 billiard balls need.

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