Question

If you deposit $100 each month into an IRA earning 2.3% interest, how much will you have in the account after 17 years? Round your answer to the nearest cent.

Answer 1

Given:

The amount deposited each month, d=$100.

The rate of interest, R=2.3%.

The number of years after which the balance in the account is calculated, N=17.

The formula for the balance in the acoount after N years is,

$P_N=\frac{d((1+\frac{r}{k})^{Nk}-1)}{(\frac{r}{k})}\text{ ---(1)}$

Here, r is the interest rate in decimal form and k is the number of compounding periods in one year.

Since deposit is made every month, we use monthly compounding, k=12.

The rate of interest in decimal form is,

$r=\frac{R}{100}=\frac{2.3}{100}=0.023$

Now, substitute the known values in equation (1).

$\begin{gathered} P_{17}=\frac{100((1+\frac{0.023}{12})^{17\times12}-1)}{(\frac{0.023}{12})}\text{ } \\ P_{17}=\frac{100((1+\frac{0.023}{12})^{204}-1)}{(\frac{0.023}{12})}\text{ } \\ P_{17}=24934.19 \end{gathered}$

Therefore, after 17 years, the balance in the account will be $24934.19, to the nearest cent.