will need 3000 for a down payment on house in 10 years. account earns 6% compounded monthly, how much needs to be deposited now to reach that goal?
A(1+.06/12)^(12*10) = 3000
Now just solve for A
To calculate the amount that needs to be deposited now to reach the goal of $3000 in 10 years, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value (the goal amount of $3000)
P = the principal amount (the initial deposit we're trying to find)
r = the annual interest rate (6% or 0.06)
n = the number of times interest is compounded per year (12 for monthly compounding)
t = the number of years (10)
Let's substitute the given values into the formula and solve for P:
3000 = P(1 + 0.06/12)^(12*10)
First, let's simplify the exponent:
3000 = P(1 + 0.005)^(120)
Next, let's calculate the value inside the parentheses:
3000 = P(1.005)^(120)
To isolate P, we divide both sides of the equation by (1.005)^120:
3000 / (1.005)^120 = P
Using a calculator, we find:
P ≈ $1,889.86
Therefore, you would need to deposit approximately $1,889.86 now to reach your goal of $3000 in 10 years, assuming the account earns 6% interest compounded monthly.