Lee Holmes deposited $16,600 in a new savings account at 9% interest compounded semiannually. At the beginning of year 4, Lee deposits an additional $41,600 at 9% interest compounded semiannually.

At the end of year 6, what is the balance in Lee’s account?
I have tried and I got $44691.78 but that is incorrect. I solved each part separately then added the what he deposited. PLEASE HELP ME!!
Math - Steve, Wednesday, October 12, 2011 at 1:58pm
It's not quite clear from the language, but it appears that after 3 years, an additional deposit was made, and then the current balance is left to draw 9% interest for two more years. If that's the case, then

to start: 16600
at the end of year 3, he has 16600*1.045^6 = 21617.52

Add 41600 = 63217.52

After two more years, 63217.52*1.045^4 = 75388.07

Math - Steve, Wednesday, October 12, 2011 at 2:00pm
Oops. That just takes us 5 years.

63217.52*1.045^6 = 82325.66

Math - Reiny, Wednesday, October 12, 2011 at 4:50pm
Make a time graph to see how the periods work

amount = 16600(1.045)^12 + 41600(1.045)^6
= 82 325.65

Math - Lydia, Saturday, October 15, 2011 at 9:08am
Thannk you both of you but both answers are incorrect. I can input the answer to see if it is right and both are wrong. We are all missing something. If anyone else wants to try please do!!!

Reiny your answer is not even close to correct.

Principal=$16,600

r=9%/2 (semiannually)=4.5%
n= 3yrs@2 (periods per year)= 6
Factor per chart= 1.3023

After 3 years p*f
16,600*1.3023= 21618.18
add deposit from yr4
21618.18+41,600.00= 63218.18
since it is the same time period (3rs) and same pecentage (9%) the factor is the same.
1.3023*63218.18=$82,329.04

To calculate the balance of Lee's account at the end of year 6, we need to consider both deposits made and the interest compounded semiannually. Let's break down the steps to solve this problem:

1. Determine the balance after 3 years (before the additional deposit):
- Initial deposit: $16,600
- Interest rate: 9% compounded semiannually
- Time: 3 years (6 semiannual periods)
- Calculate the balance after 3 years: $16,600 * (1 + 0.09/2)^6 = $21,617.52

2. Add the additional deposit made at the beginning of year 4:
- Additional deposit: $41,600
- Total balance after the additional deposit: $21,617.52 + $41,600 = $63,217.52

3. Calculate the balance after 2 more years:
- Time: 2 years (4 semiannual periods)
- Calculate the balance after 2 more years: $63,217.52 * (1 + 0.09/2)^4 = $75,388.07

Therefore, the balance in Lee's account at the end of year 6 is $75,388.07.

It seems that the previous calculations may not have accounted for the compounding periods correctly. By calculating the interest over the correct number of periods, we arrive at the correct balance.

I stand by my solution where I said"

"Math - Reiny, Wednesday, October 12, 2011 at 4:50pm
Make a time graph to see how the periods work

amount = 16600(1.045)^12 + 41600(1.045)^6
= 82 325.65 "

BTW, Steve had the same answer.

Proof my answer is correct:

rate = .045 per half-year

Balance now: 16600
balance after 1 yr = 16600(1.045)^2 = 18127.62
balance after 2 yrs. = 18127.62(1.045)^2 = 19795.81
balance after 3yrs = 19795.81(1.045)^2 = 21.617.52
at that point, beginning of year 4, an additional 41600 in added
balance after 3 yrs = 63217.52
balance after 4 yrs = 63217.52(1.045)^2 = 69035.11
balance after 5 yrs = 69035.11(1.045)^2 = 75388.07
balance after 6 yrs = 75388.07(1.045)^2 = 82325.65

PLease accept the inescapable conclusion that this answer is correct