SOLUTION: A principal of ​$7000 is invested in an account paying an annual rate of 4​%. Find the amount in the account after 3 years if the account is compounded​ semiannua
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Question 1090542: A principal of $7000 is invested in an account paying an annual rate of 4%. Find the amount in the account after 3 years if the account is compounded semiannually, quarterly, and monthly.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the formula to use is:
f = p * (1 + r) ^ n
f is the future value
p is the present value
r is the interest rate per time period
n is the number of time periods.
the interest rate per time period is determined as follows.
the interest rate per semi-annual time period is equal to the interest rate per year divided by 2.
the interest rate per quarterly time period is equal to the interest rate per year divided by 4.
the interest rate per monthly time period is equal to the interest rate per year divided by 12.
the number of time periods is determined as follows:
the number of semi-annual time periods is equal to the number of years times 2.
the number of quarterly time periods is equal to the number of years times 4.
the number of monthly time periods is equal to the number of years times 12.
interest rate is equal to interest rate percent divided by 100.
in your problem, the interest rate percent, otherwise known as the apr, is equal to 4%.
divide that by 100 to get an annual interest rate of .04.
the interest rate per year is equal to .04.
the interest rate per semi-annual time periods is equal to .04/2
the interest rate per quarterly time periods is equal to .04/4
the interest rate per monthly time periods is equal to .04/12
the number of yearly time periods is equal to 3.
the number of semi-annual time periods is equal to 2*3
the number of quarterly time periods is equal to 4*3
the number of monthly time periods is equal to 12*3
your equation of f = p * (1 + r) ^ n is set up as follows:
for annual time periods:
p = 7000
r = .04
n = 3
f = 7000 * (1 + .04) ^ 3 = 7874.048
for semi-annual time periods:
p = 7000
r = .04/2
n = 3*2
f = 7000 * (1 + .04/2) ^ (3*2) = 7883.136935
for quarterly time periods:
p = 7000
r = .04/4
n = 3*4
f = 7000 * (1 + .04/4) ^ (3*4) = 7887.775211
for monthly time periods:L
p = 7000
r = .04/12
n = 3*12
f = 7000 * (1 + .04/12) ^ (3*12) = 7890.903122
the more time periods per year, the greater the future value will be, up to the point where you have continuous compounding.
for example:
with daily compounding, the number of time periods per year is 365.
p = 7000
r = .04 / 365
n = 3 * 365
f = 7000 * (1 + .04/365) ^ (3*365) = 7892.426069
the continuous compounding formula is f = p * e^(r * n)
f is the future value
p is the present value
r is the interest rate per time period
n is the number of time periods.
when p = 7000 and r = .04 and n = 3, the future value will be equal to 7000 * e^(.04 * 3) = 7892.477961
with continuous compounding, the number of time periods per year becomes irrelevant, since all time periods per year will give you the same answer.
for example:
with 365 time periods a year, the continuous compounding formula becomes f = 7000 * e^(.04 / 365 * 3 * 365) which is equal to 7892.477962.
this is because .04 / 365 * 3 * 365 becomes .04 * 3 * 365 / 365 which becomes .04 * 3.
with discrete compounding, which is the f = p * (1 + r) ^ n formula, 365 time periods a year gets you pretty close to what the continuous compounding formula gives you.
the future value curve for discrete compounding flattens as you approach continuous compounding and can go no further than that.
this means that the incremental change gets less and less, the more time periods you use.
1000 time periods a year gets you closer to continuous compounding.
10000 time periods a year gets you closer still.
the limit is continuous compounding.
your future value can't get any higher than that.
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