Chapter 8.1, Problem 121E

(a)

To determine

To compute: the first eight terms of this sequence.

Answer to Problem 121E

The first eight terms of the sequence $A_n=5000{\left(1+\frac{0.03}{4}\right)}^n$ are given by

$A_n=5000{\left(1+\frac{0.03}{4}\right)}^n$   $A_2=5075.28$   $A_3=5113.35$   $A_4=5151.70$   $A_5=5190.33$   $A_6=5229.26$   $A_7=5268.48$   $A_8=5307.99$

Explanation of Solution

Given information:

A deposit of $5000 is made in an account that earns 3% interest compounded quarterly. The balance in the account after $n$ quarters is given by

  $A_n=5000{\left(1+\frac{0.03}{4}\right)}^n,n=1,2,3,\mathrm{...}$

Calculation:

Determine the first eight terms of the sequence $A_n=5000{\left(1+\frac{0.03}{4}\right)}^n$ as shown below:

$n$	$A_n=5000{\left(1+\frac{0.03}{4}\right)}^n$
1	$A_1=5000{\left(1+\frac{0.03}{4}\right)}^1=5000{\left(\frac{4.03}{4}\right)}^1=5037.50$
2	$A_2=5075.28$
3	$A_3=5113.35$
4	$A_4=5151.70$
5	$A_5=5190.33$
6	$A_6=5229.26$
7	$A_7=5268.48$
8	$A_8=5307.99$

Result: thus, the first eight terms of the sequence $A_n=5000{\left(1+\frac{0.03}{4}\right)}^n$ are given by

$A_n=5000{\left(1+\frac{0.03}{4}\right)}^n$   $A_2=5075.28$   $A_3=5113.35$   $A_4=5151.70$   $A_5=5190.33$   $A_6=5229.26$   $A_7=5268.48$   $A_8=5307.99$

(b)

To determine

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

Answer to Problem 121E

The balance after 10 years in this account will be $6741.74.

Explanation of Solution

Given information:

A deposit of $5000 is made in an account that earns 3% interest compounded quarterly. The balance in the account after $n$ quarters is given by

  $A_n=5000{\left(1+\frac{0.03}{4}\right)}^n,n=1,2,3,\mathrm{...}$

Calculation:

Determine the balance in this account after 10 years by computing the 40^th term of the sequence as shown below:

Consider the sequence.

  $A_n=5000{\left(1+\frac{0.03}{4}\right)}^n$

Plug in $n=40$ to the equation

  $\begin{array}{c}{A}_{40}=5000{\left(1+\frac{0.03}{4}\right)}^{40}\\ =5000{\left(\frac{4.03}{4}\right)}^{40}\\ =5000\left(1.348348612233641\right)\\ =6741.74306168205\\ =6741.74\end{array}$

Result: Thus, the balance after 10 years in this account will be $6741.74.