Chapter 8, Problem 25RE

a.

To determine

To write: the first eight terms of the sequence.

Answer to Problem 25RE

  $\begin{array}{c}{A}_1\approx \$3015\\ A_2\approx \$3030.06\\ A_3\approx \$3045.21\\ A_4\approx \$3060.45\end{array}$

And

  $\begin{array}{c}{A}_5\approx \$3075.75\\ A_6\approx \$3091.12\\ A_7\approx \$3106.56\\ A_8\approx \$3122.07\end{array}$

Explanation of Solution

Given:

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

  $A_n=3000{\left(1+\frac{0.02}{4}\right)}^n$ , $n=1,2,3,\mathrm{...}$

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

  $\begin{array}{c}{A}_1=3000{\left(1+\frac{0.02}{4}\right)}^1\\ \approx \$3015\\ A_2=3000{\left(1+\frac{0.02}{4}\right)}^2\\ \approx \$3030.06\\ A_3=3000{\left(1+\frac{0.02}{4}\right)}^3\\ \approx \$3045.21\\ A_4=3000{\left(1+\frac{0.02}{4}\right)}^4\\ \approx \$3060.45\end{array}$

And

  $\begin{array}{c}{A}_5=3000{\left(1+\frac{0.02}{4}\right)}^5\\ \approx \$3075.75\\ A_6=3000{\left(1+\frac{0.02}{4}\right)}^6\\ \approx \$3091.12\\ A_7=3000{\left(1+\frac{0.02}{4}\right)}^7\\ \approx \$3106.56\\ A_8=3000{\left(1+\frac{0.02}{4}\right)}^8\\ \approx \$3122.07\end{array}$

b.

To determine

To find: the balance in the account after 10 years by computing the 40th term of the sequence.

Answer to Problem 25RE

  $\$3662.37$

Explanation of Solution

Given:

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

  $A_n=3000{\left(1+\frac{0.02}{4}\right)}^n$ , $n=1,2,3,\mathrm{...}$

To find the balance in the account after 10 years substitute $n=40$ in the above sequence.

  $\begin{array}{c}{A}_{40}=3000{\left(1+\frac{0.02}{4}\right)}^{40}\\ =3000{\left(1.005\right)}^{40}\\ \approx 3662.37\end{array}$