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The sequence is:

$$4,\frac{4a}{3},\frac{4a^2}{9},\frac{4a^3}{27}$$

My formula is far is $a_n~=~4\cdot \frac{a^{n-1}}{3}$

Math is my worst subject, I would appreciate help or even guidance on how to do this. My math teacher is away.

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    $\begingroup$ Hint: first term is $4$ and common ratio is $\dfrac{4a}{3} \Big/ 4\,$. $\endgroup$ – dxiv Jul 23 '18 at 18:04
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    $\begingroup$ Thank you so much for your time. I really do appreciate every comment/hint, have a great day. :) $\endgroup$ – Bill Jul 23 '18 at 18:15
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Each term is $a/3$ times the previous one, so a constant $k$ exists with $a_n = k(a/3)^{n-1}$. Any one value of $n$ will let you work out $k$. Assuming $a_1=4$ instead of $a_0=4$ (sequences sometimes start at $n=0$ if it's convenient), $a_n=4(a/3)^{n-1}$, which isn't that far from what you tried.

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  • $\begingroup$ Thank you! Means a lot, enjoy the rest of your day. $\endgroup$ – Bill Jul 23 '18 at 18:15
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First of all I guess it is a little bit messy to name your sequence $a_n$ and your variable within it $a$.

Secondly I would suggest to just compute the terms to verify your solution. The given formula would lead you to

$$\frac{4}{3},\frac{4a}{3}, \frac{4a^2}{3},\frac{4a^3}{3}$$

and I guess you can identify your mistake by yourself. Also try to make clear whether you are starting with $a_0$ or with $a_1$ because this can conduct in some major problems.

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  • $\begingroup$ Thank you for taking the time out of your day to help me and others, I really appreciate it. $\endgroup$ – Bill Jul 23 '18 at 18:14

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