My Math Forum Very challenging sequence 1

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April 29th, 2015, 02:15 PM   #1
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Very challenging sequence 1

Hi! Does anyone know how to solve this problem! Thanx
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 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 $r$. Thanks from matisolla
 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
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Quote:
 Originally Posted by greg1313 I believe that's "2x", not "zx". matisolla, what have you tried?

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!
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 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 $\displaystyle \frac{a(r^x-1)}{r-1}=200\Rightarrow r-1=\frac{a(r^x-1)}{200}$ $\displaystyle \frac{a(r^{2x}-1)}{r-1}=\frac{200a(r^{2x}-1)}{a(r^x-1)}=16400$ $\displaystyle \frac{r^{2x}-1}{r^x-1}=82$ $\displaystyle \frac{(r^x-1)(r^x+1)}{r^x-1}=82$ $\displaystyle r^x+1=82$ $\displaystyle r^x=81$ $\displaystyle a=5,r=3\text{ (with }x=4)\text{ or }a=20,r=9\text{ (with }x=2).$ Your handwriting is difficult to read. Type out your posts!

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