Chapter 8.1, Problem 122E

(a)

To determine

To write: the first eight terms of the sequence.

Answer to Problem 122E

The first 8 terms of the sequence are:

10029.17, 10058.42, 10057.76, 10117.18, 101746.69, 10176.28, 10205.96, 10235.73.

Explanation of Solution

Given information:

A deposit of $10000 is made in an account that earns 3.5% interest compounded quarterly. The balance in the account after $n$ quarters is given by

  $A_n=10000{\left(1+\frac{0.035}{12}\right)}^n,n=1,2,3,\mathrm{...}$

Calculation:

Consider that the balance in the account after $n$ months is:

  $A_n=10000{\left(1+\frac{0.035}{12}\right)}^n$

To find the first term, substitute $n=1$ in the sequence.

  $\begin{array}{c}{A}_n=10000{\left(1+\frac{0.035}{12}\right)}^n\\ =10029.17\end{array}$

To find the next seven terms substitute $n=2,n=3,n=4,n=5,n=6,n=7,n=8$ in the sequence one by one.

  $\begin{array}{c}{A}_2=10000{\left(1+\frac{0.035}{12}\right)}^2\\ =10058.42\\ A_3=10057.76\\ A_4=10117.18\\ A_5=10146.69\\ A_6=10176.28\\ A_7=10205.96\\ A_8=10235.73\end{array}$

Thus, the first 8 terms of the sequence are:

10029.17, 10058.42, 10057.76, 10117.18, 101746.69, 10176.28, 10205.96, 10235.73.

(b)

To determine

To find: the balance in the account after 5 years by computing the 60^th term of the sequence.

Answer to Problem 122E

The balance in the account after 5 years is $11909.43.

Explanation of Solution

Given information:

A deposit of $10000 is made in an account that earns 3.5% interest compounded quarterly. The balance in the account after $n$ quarters is given by

  $A_n=10000{\left(1+\frac{0.035}{12}\right)}^n,n=1,2,3,\mathrm{...}$

Calculation:

The 60^th term is:

  $\begin{array}{c}{A}_{60}=10000{\left(1+\frac{0.035}{12}\right)}^{60}\\ =11909.43\end{array}$

Thus, the balance in the account after 5 years is $11909.43.

(c)

To determine

To explain: whether the balance after 10 years twice the balance after 5 years.

Answer to Problem 122E

The balance in the account after 10 years is not equal to twice the balance after 5 years.

Explanation of Solution

Given information:

A deposit of $10000 is made in an account that earns 3.5% interest compounded quarterly. The balance in the account after $n$ quarters is given by

  $A_n=10000{\left(1+\frac{0.035}{12}\right)}^n,n=1,2,3,\mathrm{...}$

Calculation:

To find the balance after 10 years , find the ${120}^{th}$ term,

  $\begin{array}{c}{A}_{120}=10000{\left(1+\frac{0.035}{12}\right)}^{120}\\ =14183.45\end{array}$

Thus,

  $A_{120}\ne 2\cdot A_{120}$

Therefore, the balance in the account after 10 years is not equal to twice the balance after 5 years.