The question that I am struggling with is as follows:

On the day of her birth, 1st January 1998, Mary's grandparents invested $x in a savings account. They continued to deposit $x on the first day of each month thereafter. The account paid a fixed rate of 0.4% interest per month. The interest was calculated on the last day of each month and added to the account. Let $A_n be the amount in Mary's account on the last day of the nth month, immediately after the interest had been added.

As soon as Mary was 18 she decided to invest $15000 of thismoney in an account of the same type earning 0.4% interest per month. She withdraws $1000 every year on her birthday to buy herself a present. Determine how long it will take until there is no money in the account.

There are 5 questions (Question A to E) related to thequestion described in the first paragraph above. I was able to answer and understand all the questions apart from the last one, which is described in the second paragraph above.

The mark scheme says:

r = 1.004^12 (= 1.049...)

15000r^n - [1000(r^n-1)]/(r-1) = 0

n = 27.8...

So n = 28

My understanding of this working above is that "r"is the common ratio of the geometric sequence for 1 year, "15000r^n"gives us the amount of interest plus principal you can gain over "n" years, and "[1000(r^n-1)]/(r-1)" gives us the total amount that you withdraw over the same period.

Correct?

However, I do not understand why the total amount forwithdrawal is "[1000(r^n-1)]/(r-1)" because the question says "She withdraws $1000 every year on her birthday...".So I do not understand why it is not simply"1000n" because it is a fixed amount of $1000 annually. Obviously "[1000(r^n-1)]/(r-1)" is the sum of geometric sequence with the first term of 1000 and common ratio of r, which is 1.004^12 (= 1.049...), however, no part of the question seems to even suggest that the amount of withdrawal has to increase like this each year.

I would much appreciate if someone can help me to understand this working on this mark scheme.

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