1. ## confused

Bob deposited $25k in a savings account @ 10% interest compounded semiannually. At the beginning of year 4 he deposited another$40k @10% compounded semiannually. At the end of 6yrs, what is the balance in his account? I have worked it out several ways, but now I am unsure of which way to go....

3yrsx2=6 periods
10% /2=5%

25000x1.7716 (per table in my book)=44290-25000=$19290 interest for 4yrs off$25000

2yrsx2=4 periods

40000x1.4641=58564-40000=18564 interest for 2 yrs off $40000 25000+40000+19290+18564=102,854 I also worked it out this way....... 4yrsx2=8 periods; 5% 25000x2.1436(book table)=53590-25000=$28590 interest for 4 yrs off $25000 40000x1.4641=5864-40000=18564 interest for 2yrs off$40000

25000+40000+28590+18564=$112154 2. ## Re: confused really? ... does$19290 interest on a 3 year investment of $25000 at 10% make sense? if so, let me know where that institution is ... I'll deposit all I have ASAP. general equation for compounded interest is $A = P\left(1 + \frac{r}{n}\right)^{nt}$ where A = account balance after t years paid at an APR of r, compounded n times a year. after 3 years ... $A = 25000(1.05)^6 = 33502.39$ add 40000 at the end of year 3 ... $A = 73502.39(1.05)^6 = 98500.23$ at the end of 6 years 3. ## Re: confused You could also do this by calculating the balance on the first amount for the full 6 years: $25000(1+ .05)^{12}= (25000)(1.7958)= 44896.41$ the the balance on the second amount for three years: $40000(1.05)^{6}= (40000)(1.3401)= 53603.82$ and then adding:$98500.23.

The powers of "12" and "6" are, of course, the 12 "half years" in 6 years and 6 "half years" in 3 years. And, of course, the "1.05" is the 10% per year divided by 2 since we are dealing with half years, together with the "1" for the original amount.