Question

Red balls are arranged in rows to form an equilateral triangle. The first row consists of one ball the second of two and so on upto n rows. If 669 more balls are added then all the balls can be arranged in the shape of a square and each of its sides then contains 8 balls less than each side of the triangle. Find the initial number of balls.

Answer 1

Solution :-

The formula for the sum of infinite arithmetic progression, such as the total number of balls in the triangle, is 

S(n) = n[A(1) + A(n)]/2

A(1) is the sum of first term (or top row in triangle) 

A(n) is the sum of the nth term (or edge line in triangle)

A(n) is also the value of the side of the triangle.

Here, A(1) = 1

A(n) = n

Sn = n(1 + n)/2

Taking n = x

(x - 8)² = x(x + 1)/2 + 669

⇒ x² 16x + 64 = (x² + x)/2 + 669

Multiplying it by 2, we get

⇒ 2x² - 32x + 128 = x² + x + 1338

⇒ 2x² - x² - 32x - x + 128 - 1338 = 0

⇒ x² - 33x - 1210 = 0

⇒ x² - 55x + 22x - 1210 = 0

⇒ x(x - 55) + 22x(x - 55) = 0

⇒ (x - 55) (x + 22) = 0

(x - 55) (x + 22) is the solution.

Where x > 0, x = 55

S(n) = 55(55 + 1)/2 

⇒ (55*56)/2

⇒ 3080/2

= 1540

So, initially there were 1540 balls.

Answer 2

Balls in first row = 1
Balls in second row = 2
Difference between both rows is the same. 
Thus supposing they are in the order 1,2,3
In the 'nth' row there will be 'n' number of balls
so

Sn= n÷2 [2a+(n-1)×a]
=n÷2 [2×1+ (n-1)×1]
=n÷2(2+n-1)
= n÷2 ×(n+1)
=n(n+1)÷2
 
Balls making a square
=(n-8)(n-8)=(n-8)²

n(n+1)÷2 +669 = (n-8)²
n(n+1)+1338 = 2 (n²- 16n + 64)
n²+n+1338 = 2n² - 32n +128
n² - 33n -1210 = 0
(n - 55) (n + 22) = 0
n - 55 =0  or n + 22 =0
n = 55 or n = -22

Number of rows cannot be negative thus there would be 55 rows in total. 

And the total number of balls from the beginning would be 
 Sn = 55 (55+1) ÷ 2
55 × 56 ÷ 2
Thus the result is 1540.