Chapter 4, Problem 20RE

### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

Chapter
Section

### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
1 views

# Finding Account Balances In Exercises 19-22, complete the table to determine the balance A for P dollars invested at rate r for t years, compounded n times per year. n 1 2 4 12 365 Continuous compounding A P = $7000 , r = 6 % , t = 20 years To determine To calculate: The amount A for corresponding value of n that is, number of times compounded per year in the following table when$7000 is invested at the rate of 6% for 20 years:

 n 1 2 4 12 365 Continuous compounding A
Explanation

Given Information:

The amount of $7000 is invested at the rate of 6% for 20 years and the table is as follows:  n 1 2 4 12 365 Continuous compounding A Formula used: The formula to compute the balance amount after t years when interest is compounded n times per year is, A=P(1+rn)nt Where, P is the deposited amount, A is the amount after t years, r is the interest rate in decimals, t is the number of years and n is the number of times compounded per year. The formula to compute the balance amount after t years when interest is compounded continuously is, A=Pert Where, P is the deposited amount, A is the amount after t years, r is the interest rate in decimals and t is the number of years. Calculation: Consider that the amount of$7000 is invested at the rate of 6% for 20 years.

Here, P=$7000, r=6% and t=20 years. Simplify the rate as, r=6%=6100=0.06 When n=1, Substitute 1 for number of times compounded per year, 0.06 for interest rate,$7000 for the amount deposited and 20 years for the time in the formula, “A=P(1+rn)nt” as,

A=P(1+rn)nt=$7000(1+0.061)(1)(20) Solve further as, A=$7000(1.06)20$70003.207136$22,449.95

When n=2,

Substitute 2 for number of times compounded per year, 0.06 for interest rate, $7000 for the amount deposited and 20 years for the time in the formula, “A=P(1+rn)nt” as, A=P(1+rn)nt=$7000(1+0.062)(2)(20)

Solve further as,

A=$7000(1.03)40$70003.262038$22,834.26 When n=4, Substitute 4 for number of times compounded per year, 0.06 for interest rate,$7000 for the amount deposited and 20 years for the time in the formula, “A=P(1+rn)nt” as,

A=P(1+rn)nt=\$7000(1+0

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