Chapter 9, Problem 19RE

a.

To determine

To write: the first ten terms of the sequence.

Answer to Problem 19RE

  $A_1\approx \$10,018.75$

  $A_2\approx \$10,037.54$

  $A_3\approx \$10,056.36$

  $A_4\approx \$10,075.21$

  $A_5\approx \$10,094.10$

  $A_6\approx \$10,113.03$

  $A_7\approx \$10,131.99$

  $A_8\approx \$10,150.99$

  $A_9\approx \$10,170.02$

  $A_{10}\approx \$10,189.09$

Explanation of Solution

Given:

The balance in the account after n months is given by:

  $A_n=10,000{\left(1+\frac{0.0225}{12}\right)}^n$ , $n=1,2,3,\mathrm{...}$

To evaluate the first eight terms of the sequence substitute $n=1,2,3,\mathrm{...},8$ in the above sequence.

  $\begin{array}{c}{A}_1=10,000{\left(1+\frac{0.0225}{12}\right)}^1\\ \approx \$10,018.75\end{array}$

  $\begin{array}{c}{A}_2=10,000{\left(1+\frac{0.0225}{12}\right)}^2\\ \approx \$10,037.54\end{array}$

  $\begin{array}{c}{A}_3=10,000{\left(1+\frac{0.0225}{12}\right)}^3\\ \approx \$10,056.36\end{array}$

  $\begin{array}{c}{A}_4=10,000{\left(1+\frac{0.0225}{12}\right)}^4\\ \approx \$10,075.21\end{array}$

  $\begin{array}{c}{A}_5=10,000{\left(1+\frac{0.0225}{12}\right)}^5\\ \approx \$10,094.10\end{array}$

  $\begin{array}{c}{A}_6=10,000{\left(1+\frac{0.0225}{12}\right)}^6\\ \approx \$10,113.03\end{array}$

  $\begin{array}{c}{A}_7=10,000{\left(1+\frac{0.0225}{12}\right)}^7\\ \approx \$10,131.99\end{array}$

  $\begin{array}{c}{A}_8=10,000{\left(1+\frac{0.0225}{12}\right)}^8\\ \approx \$10,150.99\end{array}$

  $\begin{array}{c}{A}_9=10,000{\left(1+\frac{0.0225}{12}\right)}^9\\ \approx \$10,170.02\end{array}$

  $\begin{array}{c}{A}_{10}=10,000{\left(1+\frac{0.0225}{12}\right)}^{10}\\ \approx \$10,189.09\end{array}$

b.

To determine

To find: the balance in the account after 10 years by computing the 120th term of the sequence.

Answer to Problem 19RE

  $\$12,520.59$

Explanation of Solution

Given:

The balance in the account after n quarters is given by:

  $A_n=10,000{\left(1+\frac{0.0225}{12}\right)}^n$ , $n=1,2,3,\mathrm{...}$

To find the balance in the account after 10 years substitute $n=120$ in the above sequence.

  $\begin{array}{c}{A}_{120}=10,000{\left(1+\frac{0.0225}{12}\right)}^{120}\\ \approx \$12,520.59\end{array}$