Question

4. Meagan invests $1,200 each year in an IRA for 12 years in an account that earned 5%

Answer 1

Part A)

$\bf \qquad \qquad \textit{Future Value of an ordinary annuity}\\ \left. \qquad \qquad \right.(\textit{payments at the end of the period}) \\\\ A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]$

$\bf \qquad \begin{cases} A= \begin{array}{llll} \textit{accumulated amount}\\ \end{array} \begin{array}{llll} \end{array}\\ pymnt=\textit{periodic payments}\to &1200\\ r=rate\to 5\%\to \frac{5}{100}\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &12 \end{cases}$

$\bf A=1200\left[ \cfrac{\left( 1+\frac{0.05}{1} \right)^{1\cdot 12}-1}{\frac{0.05}{1}} \right]\implies A\approx 19100.55$

part B)

so, for the next 11 years, she didn't make any deposits on it and simple let it sit and collect interest, compounded annually at 5%.

$\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \ \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\ 19100.55\\ r=rate\to 5\%\to \frac{5}{100}\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &11 \end{cases} \\\\\\ A=19100.55\left(1+\frac{0.05}{1}\right)^{1\cdot 11}\implies A\approx 32668.42$

part C)

well, for 12 years she deposited 1200 bucks, that means 12 * 1200, or 14,400.

now, here she is, 12+11, or 23 years later, and she's got 32,668.42 bucks?

all that came out of her pocket was 14,400, so 32,668.42 - 14,400, is how much she earned in interest.