1. ## formula for interest

Q: At the beginning of the year, a man deposited $100,000 with a bank that pays 10% interest per annum at the end of each year. After the interest is credited, he immediately withdraws$12,000. Likewise, he will again withdraw $12,000 at the end of each subsequent year, immediately after the bank's interest has been credited. After his nth withdrawal, he noticed, for the first time, that his bank account balance falls below$20,00. Find n.

2. Your basic accumulated value problem.

i = 0.10
Withdrawal Annuity is "Immediate".

$100000*(1+i)^{n} - 12000*\frac{(1+i)^{n}-1}{i} = 20000$

Convince yourself.

$100000*(1+i)^{1} - 12000*\frac{(1+i)^{1}-1}{i} = 98000$

That seems good.

$100000*(1+i)^{2} - 12000*\frac{(1+i)^{2}-1}{i} = 95800$

I think we're on the right track.

You're only remaining dilemma is which value around the exact value for 'n' will actually answer the question - the one before or the one after?

3. Hello TKHunny! Thank you for replying! May I know how did you arrive at the "general formula" in the first place?

4. Hello, Tangera!

Are we expected to derive the necessary formula?
. . It is a Herculean task . . .

At the beginning of the year, a man deposited $100,000 with a bank that pays 10% interest per annum at the end of each year. After the interest is credited, he immediately withdraws$12,000.
Likewise, he will again withdraw $12,000 at the end of each subsequent year, immediately after the bank's interest has been credited. After his $n^{th}$ withdrawal, he noticed, for the first time, that his bank account balance falls below$20,000.
Find $n.$
He deposits $P$ dollars in the account.

At the end of year 1, it has grown to: $(1.1)P$ dollars.
He withdraws $D$ dollars.
His balance is: . $(1.1)P - D$ dollars.

At the end of year 2, it has grown to: $1.1\bigg[(1.1)P - D\bigg]$ dollars.
He withdraws $D$ dollars.
His balance is: . $(1.1^2)P - (1.1)D - D$ dollars.

At the end of year 3, it has grown to: $1.1\bigg[(1.1^2)P - (1.1)D - D\bigg]$ dollars.
He withdraws $D$ dollars.
His balance is: . $(1.1^3)P - (1.1^2)D - (1.1)D - D$ dollars.

And so on . . .

At the end of year $n$ his balance is:

. . $A \;=\;(1.1^n)P - (1.1^{n-1})D -(1.1^{n-2})D - \hdots - (1.1)D - D$

. . $A \;=\;(1.1^n)P - \underbrace{\bigg[1.1^{n-1} + 1.1^{n-2} + \hdots + 1.1^2 + 1.1 + 1\bigg]}_{\text{geometric series}}D\;\;\;{\color{blue}[1]}$

. . The geometric series has: first term $a = 1$, common ratio $r = 1.1$, and $n$ terms.

. . Its sum is: . $1\cdot\frac{1.1^n-1}{1.1-1} \:=\:10(1.1^n -1)$

Then [1] becomes: . $A \;=\;(1.1^n)P - 10(1.1^n-1)D$

We want to know when $A \leq 20,000$

We have: . $(1.1^n)P - 10(1.1^n-1)D \:\leq \:20,000\quad\hdots$ and we will solve for $n.$

. . $(1.1^n)P - 10(1,1^n)D + 10D \:\leq \:20,000$

. . . . . . . . . $(1.1^n)(P - 10D) \:\leq \:20,000 - 10D$

. . . . . . . . . . . . . . . . . $1.1^n \;\;{\color{red}\geq} \;\;\frac{20,000 - 10D}{P - 10D} \quad\hdots$ Note that $P-10D$ is negative.

Take logs: . $\ln(1.1^n) \;=\;\ln\left(\frac{20,000-10D}{P - 10D}\right) \quad\Rightarrow\quad n\!\cdot\!\ln(1.1) \;\geq\;\ln\left(\frac{20,000-10D}{P-10D}\right)$

. . $\text{Hence: }\;\boxed{n \;\geq \;\frac{\ln\left(\dfrac{20,000-10D}{P-10D}\right)}{\ln(1.1)}}\quad\hdots\quad Whew!$

Let $P = 100,000\text{ and }D = 12,000$

. . and we get: . $n \:\geq \:\frac{\ln(5)}{\ln(1.1)} \;=\;16.8863... \;\approx\;{\bf 17\text{ years}}$

5. Hello Soroban! Thank you very much for helping too!
I meant to ask how can I find out what the geometric or arithmetic series is, because I am usually unable to find what is the general form of the series like [i.e. what form will the nth term take.]...Do I have to trial and error?? >.<

Thank you!

6. There are formulas for that sort of thing, and not very many. Soroban provided a few hints concerning their derivation.