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April 29th, 2015, 02:15 PM  #1 
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  Very challenging sequence 1
Hi! Does anyone know how to solve this problem! Thanx

April 29th, 2015, 03:16 PM  #2 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
The partial sum of a geometric series from the first term to the $n$th term is $S_n = \dfrac{a(r^n  1)}{r  1}$ Where $a$ is the first term and $r$ is the common ratio. In your question, you would end up with the formulae $\dfrac{a(r^x  1)}{r  1} = 200$ and $\dfrac{a(r^{zx}  1)}{r  1} = 16400$ Now solve for $a$ and $r$. 
April 29th, 2015, 03:50 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond 
I believe that's "2x", not "zx". matisolla, what have you tried? 
April 29th, 2015, 04:21 PM  #4  
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  Quote:
Indeed my friend, it is 2X. Sorry for the resemblance in of z and 2 in my afwul calligraphy! I attach a picture of my progress so far. I dont know how to continue from there. I would appreciate your help, since you always help me in this page!  
April 29th, 2015, 04:35 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond 
$\displaystyle \frac{a(r^x1)}{r1}=200\Rightarrow r1=\frac{a(r^x1)}{200}$ $\displaystyle \frac{a(r^{2x}1)}{r1}=\frac{200a(r^{2x}1)}{a(r^x1)}=16400$ $\displaystyle \frac{r^{2x}1}{r^x1}=82$ $\displaystyle \frac{(r^x1)(r^x+1)}{r^x1}=82$ $\displaystyle r^x+1=82$ $\displaystyle r^x=81$ $\displaystyle a=5,r=3\text{ (with }x=4)\text{ or }a=20,r=9\text{ (with }x=2).$ Your handwriting is difficult to read. Type out your posts! 

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