Question

A saving account pays 2% interest compounded annually. If $1,200 is deposited initially and again at the first of each year,how much money will be in the account three years after the initial deposit

Answer 1

3(1.02(1200))

3.06(1200) = 3672

You will have $3672

Answer 2

Answer:

The balance after 3 years will be $3745.93.

Step-by-step explanation:

The rate of interest = 2% or 0.02

p = 1200

t = 3

n = 1

1st year:

Compound interest formula is :

$p(1+r/n)^{nt}$

= $1200(1+0.02/1)^{3}$

= $1200(1.02)^{3}$

= $1273.45

2nd year:

$1200(1+0.02)^{2}$

= $1200(1.02)^{2}$

=$1248.48

3rd year:

$1200(1+0.02)^{1}$

= $1200(1.02)^{1}$

=$1224

So, total amount will be = $1273.45+1248.48+1224=3745.93$ dollars

The balance after 3 years will be $3745.93.