Answered: 4 12 Continuous Compounding 1 365

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter5: Linear Inequalities

Section5.2: Solving Inequalities By Multiplication And Division

Problem 3AGP

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Question

Complete the table by determining the balance A for P dollars invested at rate r for t years and compounded n times per year P = $2500 r = 6% t = 20 years

Expert Solution

Step 1

Given: P=$2500, r=6%, t=20 years

The formula for Compound Interest is:

$A = P {(1 + \frac{\frac{r}{n}}{100})}^{t n}$ , where A is the amount, P is the principal, r is the rate of interest, n is the number of times compounding is done in a year and t is the time period.

For n=1:

$\begin{array}{rcl} A & = & P {(1 + \frac{\frac{r}{n}}{100})}^{t n} \\ = & 2500 {(1 + \frac{6}{100})}^{20} \\ = & 2500 {(1.06)}^{20} \\ = & 8, 017.83 \end{array}$

Thus, when n=1 the value of the amount is $8,017.83

Step 2

For n=2:

$\begin{array}{rcl} A & = & P {(1 + \frac{\frac{r}{n}}{100})}^{t n} \\ = & 2500 {(1 + \frac{\frac{6}{2}}{100})}^{20 \times 2} \\ = & 2500 {(1 + \frac{3}{100})}^{40} \\ = & 2500 {(1.03)}^{40} \\ = & 8, 155.09 \end{array}$

Thus, when n=2 the value of the amount is $8,155.09.

For n=4:

$\begin{array}{rcl} A & = & P {(1 + \frac{\frac{r}{n}}{100})}^{t n} \\ = & 2500 {(1 + \frac{\frac{6}{4}}{100})}^{20 \times 4} \\ = & 2500 {(1 + \frac{1.5}{100})}^{80} \\ = & 2500 {(1.015)}^{80} \\ = & 8, 226.65 \end{array}$

Thus, when n=4 the value of the amount is $8,226.65.

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