Question

Finding Account Balances In Exercise,complete the table to determine the table A for P dollars invested at rate r for t years,compounded n times per year .see Example 3.
n 1 2 4 12 365 continuous compounding
A
p = $1000,r = 3%,t = 10 years

Answer 1

Answer:

n = 1, A = $1,343.92

n = 2, A = $1,346.86

n = 4, A = $1,348.35

n = 12, A = $1,349.35

n = 365, A = $1,349.84

Continuous compounding, A = $1,349.86

Step-by-step explanation:

We are given the following in the question:

P = $1000

r = 3% = 0.03

t = 10 years

Formula:

The compound interest is given by

$A = P\bigg(1 + \displaystyle\frac{r}{n}\bigg)^{nt}$

where P is the principal, r is the interest rate, t is the time, n is the nature of compound interest and A is the final amount.

For n = 1

$A = 1000\bigg(1 + \displaystyle\frac{0.03}{1}\bigg)^{10}\\\\A = \ 1,343.92$

For n = 2

$A = 1000\bigg(1 + \displaystyle\frac{0.03}{1}\bigg)^{20}\\\\A = \ 1,346.86$

For n = 4

$A = 1000\bigg(1 + \displaystyle\frac{0.03}{1}\bigg)^{40}\\\\A = \ 1,348.35$

For n = 12

$A = 1000\bigg(1 + \displaystyle\frac{0.03}{1}\bigg)^{120}\\\\A = \ 1,349.35$

For n = 365

$A = 1000\bigg(1 + \displaystyle\frac{0.03}{1}\bigg)^{3650}\\\\A = \ 1,349.84$

Continuous compounding:

$A = Pe^{rt}\\A = 1000e^{0.03\times 10}\\A = \ 1,349.86$