# If a payment $P is made every year into an account that attracts a fixed interest rate of r % per annum compounded continuously pans the account is closed t years later the balance due will be: ...

If a payment $P is made every year into an account that attracts a fixed interest rate of r % per annum compounded continuously pans the account is closed t years later the balance due will be:

`(P(e^0.01rt-1)/(1-e^-0.01r)` )

where e is eulers number

a.) find the balance due after 10 years if 2000 dollars is invested each year and the interest rate is fixed at 10% per annum compounded continuously

b. ) find the number if years the scheme must run if an investor wants to invest 3000 dollars per year and close the account whenthe balance reaches 154000 dollars, assuming a constant interest rate of 8 percent per year compounded continuosly

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I am assuming your expression is

`A=(p(e^(.01rt)-1))/(1-e^(-.01r))`

If it is incorrect,we will have to work out,

$2000 is invested each year, so P = 2000.

r is the interest rate, which is given as 10%. Thus, r = 10

We need to find the balance due after 10 years, so t = 10.

Hence, the balance due will be

Balance A=`(2000(e^(.01xx10xx10)-1))/(1-e^(-.01xx10))`

`=(2000xx1.7182818)/(0.09510)`

`=36135.19$`

**b.**

$3000 is invested each year, so P = 3000.

r is the interest rate, which is given as 8%. Thus, r = 8

We need to find the balance =15400

r is the interest rate, which is given as 8%. Thus, r = 8

We need to find the balance =15400

due after years, so t = ?.

`15400=(3000(e^(.01xx8xxt)-1))/(1-e^(-.01xx8))`

`(15400xx.0768836)/3000=e^(.08t)-1`

`.394669+1=e^.08t`

`1.394669=e^(.08t)`

`.08t=ln(1.394669)`

`.08t=.332657111`

`t=4.158` (approx)

approximately

t=4 years.