Chapter 9.1, Problem 97E

a.

To determine

To write: the first eight terms of the sequence.

Answer to Problem 97E

  $A_1\approx \$10,087.50$

  $A_2\approx \$10,175.77$

  $A_3\approx \$10,264.80$

  $A_4\approx \$10,354.62$

  $A_5\approx \$10,445.22$

  $A_6\approx \$10,536.62$

  $A_7\approx \$10,628.81$

  $A_8\approx \$10,721.82$

Explanation of Solution

Given:

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

  $A_n=10,000{\left(1+\frac{0.035}{4}\right)}^n$ , $n=1,2,3,\mathrm{...}$

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

  $\begin{array}{c}{A}_1=10,000{\left(1+\frac{0.035}{4}\right)}^1\\ \approx \$10,087.50\end{array}$

  $\begin{array}{c}{A}_2=10,000{\left(1+\frac{0.035}{4}\right)}^2\\ \approx \$10,175.77\end{array}$

  $\begin{array}{c}{A}_3=10,000{\left(1+\frac{0.035}{4}\right)}^3\\ \approx \$10,264.80\end{array}$

  $\begin{array}{c}{A}_4=10,000{\left(1+\frac{0.035}{4}\right)}^4\\ \approx \$10,354.62\end{array}$

  $\begin{array}{c}{A}_5=10,000{\left(1+\frac{0.035}{4}\right)}^5\\ \approx \$10,445.22\end{array}$

  $\begin{array}{c}{A}_6=10,000{\left(1+\frac{0.035}{4}\right)}^6\\ \approx \$10,536.62\end{array}$

  $\begin{array}{c}{A}_7=10,000{\left(1+\frac{0.035}{4}\right)}^7\\ \approx \$10,628.81\end{array}$

  $\begin{array}{c}{A}_8=10,000{\left(1+\frac{0.035}{4}\right)}^8\\ \approx \$10,721.82\end{array}$

b.

To determine

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

Answer to Problem 97E

  $\$14,169.09$

Explanation of Solution

Given:

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

  $A_n=10,000{\left(1+\frac{0.035}{4}\right)}^n$ , $n=1,2,3,\mathrm{...}$

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

  $\begin{array}{c}{A}_{40}=10,000{\left(1+\frac{0.035}{4}\right)}^{40}\\ \approx \$14,169.09\end{array}$

c.

To determine

Is the balance after 20 years twice the balance after 10 years.

Answer to Problem 97E

No

Explanation of Solution

Given:

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

  $A_n=10,000{\left(1+\frac{0.035}{4}\right)}^n$ , $n=1,2,3,\mathrm{...}$

To find the balance in the account after 20 years substitute $n=80$ in the above sequence.

  $\begin{array}{c}{A}_{80}=10,000{\left(1+\frac{0.035}{4}\right)}^{80}\\ \approx \$20,076.31\end{array}$

And from part (b),

  $A_{40}=\$14,169.09$

Clearly,

  $A_{80}=\$20,076.31\ne 2A_{40}$