Problem: Find the amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time. Determine whether the list of numbers involved makes an arithmetic progression, and explain why.

Solution:

To solve the problem, we can use the formula for geometric progression:

a_n = a₁ * r^n-1

where:

- a_n is the nth term of the sequence
- a₁ is the first term of the sequence
- r is the common ratio between the terms
- n is the number of terms in the sequence

Let's assume that the amount of air in the cylinder is a₁. After the first removal of 1/4 of the air, the amount of air remaining in the cylinder is 3/4 of the original amount:

a₂ = 3/4 * a₁

After the second removal of 1/4 of the remaining air, the amount of air remaining in the cylinder is 3/4 of the previous amount:

a₃ = 3/4 * a₂ = (3/4)^2 * a₁

Similarly, after the third removal of 1/4 of the remaining air, the amount of air remaining in the cylinder is 3/4 of the previous amount:

a₄ = 3/4 * a₃ = (3/4)^3 * a₁

We can see that the list of numbers involved makes a geometric progression, not an arithmetic progression. This is because the common ratio between the terms is constant:

r = 3/4

To find the amount of air remaining in the cylinder after n removals, we can use the formula for the nth term:

a_n = (3/4)^(n-1) * a₁

Therefore, the amount of air remaining in the cylinder after 3 removals is:

a₄ = (3/4)^3 * a₁ = 27/64 * a₁

In conclusion, the amount of air remaining in the cylinder after 3 removals is 27/64 of the original amount. The list of numbers involved makes a geometric progression, not an arithmetic progression, because the common ratio between the terms is constant.