Chapter 4, Problem 21RE

### 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
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# 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 = $3000 , r = 3.5 % , t = 10 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$3000 is invested at the rate of 3.5% for 10 years,

 n 1 2 4 12 365 Continuous compounding A
Explanation

Given Information:

The amount of $3000 is invested at the rate of 3.5% for 10 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$3000 is invested at the rate of 3.5% for 10 years.

Here, P=$3000, r=3.5% and t=10 years. Simplify the rate as, r=3.5%=3.5100=0.035 When n=1, Substitute 1 for number of times compounded per year, 0.035 for interest rate,$3000 for the amount deposited and 10 years for the time in the formula, A=P(1+rn)nt as,

A=P(1+rn)nt=$3000(1+0.0351)(1)(10) Solve further as, A=$3000(1.035)10$30001.41060$4231.80

When n=2,

Substitute 2 for number of times compounded per year, 0.035 for interest rate, $3000 for the amount deposited and 10 years for the time in the formula, A=P(1+rn)nt as, A=P(1+rn)nt=$3000(1+0.0352)(2)(10)

Solve further as,

A=$3000(1.0175)20$30001.414778$4244.33 When n=4, Substitute 4 for number of times compounded per year, 0.035 for interest rate,$3000 for the amount deposited and 10 years for the time in the formula, A=P(1+rn)nt as,

A=P(1+rn)nt=\$3000(1+0

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