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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
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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 $r$.
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April 29th, 2015, 03:50 PM   #3
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I believe that's "2x", not "zx".

matisolla, what have you tried?
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April 29th, 2015, 04:21 PM   #4
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Quote:
Originally Posted by greg1313 View Post
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
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$\displaystyle \frac{a(r^x-1)}{r-1}=200\Rightarrow r-1=\frac{a(r^x-1)}{200}$

$\displaystyle \frac{a(r^{2x}-1)}{r-1}=\frac{200a(r^{2x}-1)}{a(r^x-1)}=16400$

$\displaystyle \frac{r^{2x}-1}{r^x-1}=82$

$\displaystyle \frac{(r^x-1)(r^x+1)}{r^x-1}=82$

$\displaystyle r^x+1=82$

$\displaystyle r^x=81$

$\displaystyle a=5,r=3\text{ (with }x=4)\text{ or }a=20,r=9\text{ (with }x=2).$

Your handwriting is difficult to read. Type out your posts!
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