How much should be deposited in an account paying 3.5% interest, compounded quarterly, in order to have a balance of $ 6,000after 17years and 3months?
P = Po(1+r)^n = 6,000
r = (3.5/4)/100% = 0.00875 = Quarterly % rate expressed as a decimal.
n = (17yrs.*12mo/yr. + 3mo.)/3 = 69 Compounding periods.
Po(1.00875)^69 = 6000
1.824Po = 6000
Po = $3289.17
To find out how much should be deposited in an account paying 3.5% interest, compounded quarterly, in order to have a balance of $6,000 after 17 years and 3 months, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment (balance in the account)
P = the principal amount (initial deposit)
r = annual interest rate (in decimal form)
n = number of compounding periods per year
t = number of years
In this case, we have:
A = $6,000
r = 3.5% = 0.035 (decimal form)
n = 4 (compounded quarterly)
t = 17 years and 3 months = 17.25 years
Plugging these values into the formula, we have:
$6,000 = P(1 + 0.035/4)^(4*17.25)
Now we need to solve for P (the principal amount).
Dividing both sides of the equation by [(1 + 0.035/4)^(4*17.25)]:
P = $6,000 / [(1 + 0.035/4)^(4*17.25)]
Using a calculator, we can evaluate the right side of the equation to find P.
P ≈ $6,000 / 2.5945
P ≈ $2,313.12
Therefore, approximately $2,313.12 should be deposited in the account in order to have a balance of $6,000 after 17 years and 3 months.