On January 1st Kendra puts $100 into her bank account. On the first of each month after that she deposits an additional $100. Unfortunately, during the middle of the month she always has to withdraw 75% of her money to pay bills.

Answer

(a) Here's the timeline:

At the beginning of March, Kendra has $131.25. In the middle of March she withdraws 75% of that amount, leaving 25% in the bank.

Rounded to the nearest cent, her bank account holds

0.25(131.25) = $32.81

at the end of March.

(b) This question is similar to (b) of the example, but it asks about what's happening at a different point in the month.

At the end of the month, Kendra has 25% of the money she had at the beginning of the month. We'll call the amount she has left at the end of the month *L*_{n}. There are two ways to do this problem: we could use the answer to (b) of the example, or we could start from scratch. It's a good idea to understand both ways.

Using the example:

At the beginning of month *n*, Kendra has *M*_{n} dollars in her account.

At the end of month *n*, she has 25% of that amount left.

So

Starting from scratch:

Let *L*_{n} be the amount of money in Kendra's account at the end of *n* months. To get from *L*_{n – 1} to *L*_{n} she deposits $100, then keeps 25% of the resulting amount.

This means

*L*_{n} = 0.25((100 + *L*_{n – 1} ).

This tells us how to build each *L*_{n} from the previous one, so look at some values of *L*_{n}.

The value *L*_{n} is the sum of a geometric series with *r* = 0.25,

*a* = 0.25(100) = 25,

and *n* terms (since the exponents go from 1 to *n*). So

Thankfully, this is equivalent to the formula we found when we used the example instead of starting from scratch.