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April 6th, 2015, 09:50 AM  #1 
Member Joined: Mar 2015 From: KY Posts: 35 Thanks: 1  Geometric Sequence
Stuck on another one, I don't know what I'm doing wrong. I follow the formula and cannot get the right answer. 
April 6th, 2015, 10:09 AM  #2 
Member Joined: Mar 2015 From: KY Posts: 35 Thanks: 1 
The answer I keep getting that I think is right is $34,440. But it says it's wrong.

April 6th, 2015, 10:10 AM  #3 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
What formula did you follow? What calculation did you make?

April 6th, 2015, 10:37 AM  #4 
Member Joined: Mar 2015 From: KY Posts: 35 Thanks: 1 
I used this one 
April 6th, 2015, 12:28 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,806 Thanks: 716 
What values did you use for r, n, and a1?

April 6th, 2015, 02:02 PM  #6 
Member Joined: Mar 2015 From: KY Posts: 35 Thanks: 1 
I used: a1=3075 r=1.025 n=10 
April 7th, 2015, 05:17 AM  #7 
Member Joined: Mar 2015 From: KY Posts: 35 Thanks: 1 
Any more help?

April 7th, 2015, 06:52 AM  #8 
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 
did you try the hint that was given? $\displaystyle 3000(1.025)^{10}+3000(1.025)^9+3000(1.025)^8+ ... +3000(1.025) = $ $\displaystyle 3000\sum_{n=1}^{10} 1.025^n =$ $\displaystyle 3000 \cdot \frac{1.025(11.025^{10})}{11.025}$ 
April 7th, 2015, 07:13 AM  #9 
Member Joined: Mar 2015 From: KY Posts: 35 Thanks: 1 
This was getting me the same answer that I was getting: \$34,440. But when I worked it out individually for each year I got \$34,450 which is the right answer. Not sure where the issue is... rounding maybe? Last edited by skipjack; April 7th, 2015 at 01:12 PM. 
April 7th, 2015, 07:27 AM  #10 
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 
I got 34450.40 using the formula I posted ...


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