Question

find the value of the ordinary annuity at the end of the indicated time period. the payment R, frequency of deposits m, (which is the same as the frequency of compounding), annual interest rate r, and time t, are given below.Amount: $400monthly. 6% interest rate4 years. the future value of the given annuity is $___________ (round to the nearest cent)

Answer 1

We will have the following:

*First: We determine the values for:

$A=R(\frac{(1+\frac{r}{m})^n-1}{\frac{r}{m}})$

So, we would have:

$\begin{cases}r=0.06 \\ m=12 \\ n=4\cdot12\Rightarrow n=48 \\ R=400\end{cases}$

*Second: We replace in the expression:

$A=(400)(\frac{(1+\frac{0.06}{12})^{48}-1}{\frac{0.06}{12}})\Rightarrow A=21639.13289$$\Rightarrow A\approx21639.13$

So, the future value of the given annuity is approxiately $21639.13.