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:

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

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.

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?

Re: Annuity reinvestment problem

Quote:

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

Re: Annuity reinvestment problem

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