1. ## Interest accumulated

A lady deposits \$50 into a fund at the end of each quarter for 10 years. Determine the total amount of interest that she has accumulated at the end of 10 years if the fund earns 4.5% per year compounded annually for the first 5 years and 5% per year compounded semiannually for the second five years.

I'm getting confused on how you set it up to find the amount of interest. I think the total amount in the account at the end of the 10 years is: 50[(1.045)^20 - 1 / .045] + 50[(1.025^20 - 1 / .025] = 2845.81
But I dont understand how you figure out how much of that is the interest, or if you even need to caclulate the total amount at all...

2. Originally Posted by tbl9301
I think the total amount in the account at the end of the 10 years is: 50[(1.045)^20 - 1 / .045] + 50[(1.025^20 - 1 / .025] = 2845.81
Try again with these:

$
50\frac{{\left( {1 + {\textstyle{{0.045} \over 1}}} \right)^{5 \times 1} - 1}}{{\left( {1 + {\textstyle{{0.045} \over 1}}} \right)^{{\textstyle{1 \over 4}}} - 1}}\left( {1 + {\textstyle{{0.05} \over 2}}} \right)^{5 \times 2} + 50\frac{{\left( {1 + {\textstyle{{0.05} \over 2}}} \right)^{5 \times 2} - 1}}{{\left( {1 + {\textstyle{{0.05} \over 2}}} \right)^{{\textstyle{2 \over 4}}} - 1}}
$

Originally Posted by tbl9301
But I dont understand how you figure out how much of that is the interest
This is quite easy as my esteemed fellow knight-errant Sir Denis (or as he is known here as Sir Wilmer) will tell you.

3. Hmm so I had been setting up in terms of "s angle n" , ie 50*(s angle 20).045 + 50*(s angle 20).025 , which is equal to what I typed above

So does your answer correlate with those terms or is that just how you're supposed to set it up ... if that makes any sense

4. Nevermind scratch that I see where you got it ... thank you!!

5. Originally Posted by jonah
This is quite easy as my esteemed fellow knight-errant Sir Denis (or as he is known here as Sir Wilmer) will tell you.
OK, Sir Jonah: "this is quite easy" .