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Assume that each deposit is made at the end of every year.

You plan to retire in 30 years and decide to save $10,000 per year. If the interest rate is 6% compounded monthly, how much will you have in 30 years? Assume that each deposit is made at the end of every year.

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Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)

   A = P ( 1 + 0.06/12)(12*30) = 10,000*(1.005)360 = 60225.75

Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
This problem asks for the future value of an annuity with monthly deposits.
The standard formula for this amount is
FV =R*((1+i)n-1)/i,
where R is the payment per compounding period, i = the interest rate per compounding period, and n the number of compounding periods. This formula assumes that one payment is made each compounding period, which in your problem is not the case: the compounding period is a month (n=30*12), while a payment R=10,000 is made once year.
We can get an approximation if we assume instead that a payment of R=$10,000/12 is made every month.
With i=0.06/12 we get
FV =(10000/12) *((1+(0.06/12))30*12-1)/(0.06/12)
=$ 837,100
This is an overestimate, because it assumes monthly payments with monthly compounding.
We can get an underestimate if we assume annual compounding with annual payments. In this case n=30, i=0.06 and
FV=10000 * ((1+0.06)30-1)/0.06
=$ 790,580
The actual answer will lie between these two estimates.
Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
4.9 4.9 (174 lesson ratings) (174)
Since the first deposit will be made by the end of the first year, there will be 29 deposits.
The first deposit will grow to be 10,000*(1+0.06/12)12*29; the second deposit grows to be 10,000*(1+0.06/12)12*28 and so on. The last deposit will grow to 10,000*(1+0.06/12)12≈$10616.78.
We need to find the sum 10,000*(1+0.06/12)12+10,000*(1+0.06/12)12*2+…+10,000*(1+0.06/12)12*29=
So we need to sum geometrical progression with the first term (1+0.06/12)12≈1.061678 and the base 1.061678, the total number of terms being 29. 
S=q*(qN-1-1)/(q-1), where q is the base of geometrical progression, N is the number of terms.
Plug in the number to get:
Final answer is S*10000=804324.48