What amount should be invested today so that there will be $5000 in an account in 10 years, if interest is earned at a rate of 5% per annum, compounded bi-weekly?
help!
Do you have any answer choices that goes to the question?
no there are no answer choices with that question
bi-weekly to me means every two weeks
so in one year there would be 27 periods and i = .05/7 = .00185185
p(1.00185185)^(270) = 5000
solve for p
To find the amount that should be invested today, we can use the compound interest formula:
A = P(1 + r/n)^(nt)
Where:
A is the future value of the investment
P is the principal amount (the amount to be invested today)
r is the annual interest rate (expressed as a decimal)
n is the number of compounding periods per year
t is the number of years
In this case, we have:
A = $5000 (the desired future value)
r = 5% per annum = 0.05 (expressed as a decimal)
n = 26 (bi-weekly compounding, since there are 52 weeks in a year and interest is compounded every two weeks)
t = 10 years
Now we can rearrange the formula to solve for P:
P = A / (1 + r/n)^(nt)
Plugging in the given values:
P = $5000 / (1 + 0.05/26)^(26*10)
Now, we can calculate this using a calculator or spreadsheet software. Let's assume we're using a calculator:
Step 1: Divide 0.05 by 26: 0.05 / 26 = 0.001923 (approx.)
Step 2: Add 1 to the result: 1 + 0.001923 = 1.001923 (approx.)
Step 3: Multiply the result by the number of compounding periods per year: 1.001923^26 = 1.050855 (approx.)
Step 4: Multiply the result by the number of years: 1.050855^260 = 1.629876 (approx.)
Step 5: Divide $5000 by the result: $5000 / 1.629876 = $3067.93 (approx.)
Therefore, you should invest approximately $3067.93 today to have $5000 in the account after 10 years, with an interest rate of 5% per annum compounded bi-weekly.