Chapter 9, Problem 64RE

To determine

Find balance at the end of 10 years

Answer to Problem 64RE

The balance in the account after 10 years is $112,072.62

Explanation of Solution

Given:

Deposit by investor=$400

Compound interest= 3%, monthly

Investor pays $800 every month for 10 years

Account pays 3% compounded monthly

At the end of first month account balance is

  $\begin{array}{l}{A}_1=800+\frac{0.03}{12}\left(800\right)\hfill \\ =800\left(1+0.0025\right)\hfill \\ =800\left(1.0025\right)\hfill \end{array}$

At the end of second month account balance is

  $\begin{array}{l}{A}_2=\left(800+A_1\right)\left(1.0025\right)\hfill \\ =800\left(1.0025\right)+800{\left(1.0025\right)}^2\hfill \end{array}$

Thus, the balance at the end of 10 years which is equal to 120 months is

  $\begin{array}{l}{A}_{120}=800\left(1.0025\right)+800{\left(1.0025\right)}^2+800{\left(1.0025\right)}^3+\mathrm{....}+800{\left(1.0025\right)}^{120}\hfill \\ =800\left(1.0025\right)\left[1+1.0025+{1.0025}^2+\mathrm{.....}+{1.0025}^{120}\right]\hfill \\ =800\left(1.0025\right)\left[\frac{{1.0025}^{120}-1}{1.0025-1}\right]\hfill \\ \approx 112,072.62\hfill \end{array}$

Hence the balance in the account after 10 years is $112,072.62