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

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

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(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(0.03)/(1)\bigg)^(10)\\\\A = \$1,343.92`

For n = 2

`A = 1000\bigg(1 + \displaystyle(0.03)/(1)\bigg)^(20)\\\\A = \$1,346.86`

For n = 4

`A = 1000\bigg(1 + \displaystyle(0.03)/(1)\bigg)^(40)\\\\A = \$1,348.35`

For n = 12

`A = 1000\bigg(1 + \displaystyle(0.03)/(1)\bigg)^(120)\\\\A = \$1,349.35`

For n = 365

`A = 1000\bigg(1 + \displaystyle(0.03)/(1)\bigg)^(3650)\\\\A = \$1,349.84`

Continuous compounding:

`A = Pe^(rt)\\A = 1000e^(0.03* 10)\\A = \$1,349.86`