Suppose a toilet rol has an outer radius of 5cm and an inner radius of 2,5cm. Suppose the thickness of tisue paper is 0.5mm. As the toilet roll unrolls, the diameter decreases. Each turn of the roll will decrease the radius of the roll by 0.5mm=0.05cm, and release less aper.
\[2 \pi r\]
of paper.Thge first turn releases
\[2 \pi \times 5=10 \pi cm\]
Thge second turn releases
\[2 \pi \times 4.95=9.9 \pi cm\]
Thge third turn releases
\[2 \pi \times 4.9=9.8 \pi cm\]
And so on. Altogether there will be
\[n= \frac{5-2.5}{0.05}+1 = 51\]
turns.
This is an arithmetic sequence with first term \[a=10 \pi\]
and common difference \[d=-0.1 \pi\]
Use the formula
\[S_n=\frac{n}{2} (2a+(n-1)d)\]
for the sum of an arithmetic sequence.\[S_{51}=\frac{51}{2} (2 \times 10 \pi+(51-1) \times (-0.1 \pi))=382.5 \pi cm \]