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 = $2500,r = 2.5%,t = 20 years

Answer 1

n = 1, A = $4,096.54

n = 2, A = $4,109.04

n = 4, A = $4,115.39

n = 12, A = $4,119.66

n = 365, A = $4,121.73

Compounded continuously, A = $4,121.80

Explanation:

We are given the following in the question:

P = $2500

r = 2.5% = 0.025

t = 20 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 = 2500\bigg(1 + \displaystyle(0.025)/(1)\bigg)^(20)\\\\A = \$4,096.54`

For n = 2

`A = 2500\bigg(1 + \displaystyle(0.025)/(2)\bigg)^(40)\\\\A = \$4,109.04`

For n = 4

`A = 2500\bigg(1 + \displaystyle(0.025)/(4)\bigg)^(80)\\\\A = \$4,115.39`

For n = 12

`A = 2500\bigg(1 + \displaystyle(0.025)/(12)\bigg)^(240)\\\\A = \$4,119.66`

For n = 365

`A = 2500\bigg(1 + \displaystyle(0.025)/(365)\bigg)^(7300)\\\\A = \$4,121.73`

Continuous compounding:

`A =Pe^(rt)\\A = 2500e^(0.025* 20)\\A = \$4,121.80`