You hire a printer to print concert tickets. he delivers them in circular rolls labeled as 1000 tickets each. you want to check the number of tickets in each roll without counting thousands of tickets. you decide to do it by measuring the diameter of the rolls. the tickets are 3 in long and 0.28 mm thick and are rolled on a core 4 cm in diameter.

Given:
The core diameter is
D = 4 cm = 400 mm
Each ticket has length of
L = 3 in = 3*254 = 762 mm,. and thickness of
t = 0.28 mm.

In order to have 1000 tickets, the unrolled length should be 
1000 * 762 = 762,000 mm

1st wrap:
diameter = D
length = πD

2nd wrap:
diameter = D + 2t
length = π(d + 2t)

3rd wrap:
diameter = D + 2(2t)
length = π[D + 2(2t)]

...

n-th wrap:
diameter = D + (n-1)*(2t)
length = π[D + (n-1)*(2t)]
The length forms an arithmetic sequence, with
a₁ = πD = 1256.6 mm (first term)
d = 2πt = 1.7593 mm

The n-th term is
1256.6 + 1.7593(n-1) = 1254.8 + 1.7593n

The total length of n wraps is
(n/2)*(1256.6 + 1254.8 + 1.7593n)
= 1255.7n + 0.8797n²

The total length should be equal to 762,000.
Therefore
0.8797n² + 1255.7n - 762000 = 0

Solve with the quadratic formula.
n = (1/1.7594)*[-1255.7 +/- √(1255.7² + 2.6812 x 10⁶ )] = 459.14 or -1886.56
Reject negative length, so that
n = 459.14, rroundto the larger value of 460.

The number of tickets will be 1000 if the 460-th wrap (outer wrap) has a diameter of
1254.8 + 1.7593*460 = 2064.1 mm, or
 2064.1/254 = 8.13 in

The diameter of each roll should be about   8.1 in (or 2064 mm)