1. ## Annuity reinvestment problem

A man invests 1000 at the beginning of each year into a fund that pays an annual interest rate of 5.6%. The annual interest payments are deposited into a fund that pays 6.2% annually. What is his total accumulation at the end of 10 years?

I thought that the way to solve this problem was:

$total accumulation = 56\"{s}_{10,0.062}+10,000=>56(\frac{1.06^{10}-1}{0.062})+10,000=10791.29$

But the answer in the back of my book says that the the annuity is increasing, but I don't understand how.

The amount of interest for year k is 1000k(0.056)=56k. The accumulation of these payments plus interest is 56(Is)10,0.06=3,730.48. Total accumulation is 13,730.48.

Can someone explain how that answer was achieved? Thanks.

2. ## Re: Annuity reinvestment problem

The 6.2% account:
Code:
YR   DEPOSIT   INTEREST  BALANCE
1      56.00        .00    56.00
2     112.00       3.47   168.47
...
10    560.00     185.09  3730.4829.....
Good nuff?

3. ## Re: Annuity reinvestment problem

Originally Posted by downthesun01
The amount of interest for year k is 1000k(0.056)=56k. The accumulation of these payments plus interest is 56(Is)10,0.06=3,730.48. Total accumulation is 13,730.48.
u = 1st payment (56)
v = constant payment increase (56)
n = number of periods (10)
i = interest rate per period (.062)
r = 1 + i (1.062)
f = future value (?)

f = { i [u(r^n - 1) - v(n - 1)] + vr[r^(n - 1) - 1]} / i^2
f = 3730.4829.....

Above also handles problems like:
Jack opens annuity account in which he deposits $500 as 1st annual payment, then increases payments by$50 each year....u = 500, v = 50

4. ## Re: Annuity reinvestment problem

Nevermind. I wasn't thinking. I the first account has 1000 added to it every year.