Help with Geometric Sequence Formula

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up vote
1
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The sequence is:



$$4,frac4a3,frac4a^29,frac4a^327$$



My formula is far is $a_n~=~4cdot fraca^n-13$



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




sequences-and-series algebra-precalculus




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  • 1




    Hint: first term is $4$ and common ratio is $dfrac4a3 Big/ 4,$.
    – dxiv
    Jul 23 at 18:04






  • 1




    Thank you so much for your time. I really do appreciate every comment/hint, have a great day. :)
    – Bill
    Jul 23 at 18:15














up vote
1
down vote

favorite












The sequence is:



$$4,frac4a3,frac4a^29,frac4a^327$$



My formula is far is $a_n~=~4cdot fraca^n-13$



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




sequences-and-series algebra-precalculus




share|cite|improve this question

















  • 1




    Hint: first term is $4$ and common ratio is $dfrac4a3 Big/ 4,$.
    – dxiv
    Jul 23 at 18:04






  • 1




    Thank you so much for your time. I really do appreciate every comment/hint, have a great day. :)
    – Bill
    Jul 23 at 18:15












up vote
1
down vote

favorite









up vote
1
down vote

favorite











The sequence is:



$$4,frac4a3,frac4a^29,frac4a^327$$



My formula is far is $a_n~=~4cdot fraca^n-13$



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




sequences-and-series algebra-precalculus




share|cite|improve this question













The sequence is:



$$4,frac4a3,frac4a^29,frac4a^327$$



My formula is far is $a_n~=~4cdot fraca^n-13$



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





sequences-and-series algebra-precalculus





share|cite|improve this question












share|cite|improve this question




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edited Jul 23 at 17:58









mrtaurho

750219




750219









asked Jul 23 at 17:54









Bill

596




596







  • 1




    Hint: first term is $4$ and common ratio is $dfrac4a3 Big/ 4,$.
    – dxiv
    Jul 23 at 18:04






  • 1




    Thank you so much for your time. I really do appreciate every comment/hint, have a great day. :)
    – Bill
    Jul 23 at 18:15












  • 1




    Hint: first term is $4$ and common ratio is $dfrac4a3 Big/ 4,$.
    – dxiv
    Jul 23 at 18:04






  • 1




    Thank you so much for your time. I really do appreciate every comment/hint, have a great day. :)
    – Bill
    Jul 23 at 18:15







1




1




Hint: first term is $4$ and common ratio is $dfrac4a3 Big/ 4,$.
– dxiv
Jul 23 at 18:04




Hint: first term is $4$ and common ratio is $dfrac4a3 Big/ 4,$.
– dxiv
Jul 23 at 18:04




1




1




Thank you so much for your time. I really do appreciate every comment/hint, have a great day. :)
– Bill
Jul 23 at 18:15




Thank you so much for your time. I really do appreciate every comment/hint, have a great day. :)
– Bill
Jul 23 at 18:15










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










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.






share|cite|improve this answer





















  • Thank you! Means a lot, enjoy the rest of your day.
    – Bill
    Jul 23 at 18:15

















up vote
2
down vote













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



$$frac43,frac4a3, frac4a^23,frac4a^33$$



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.






share|cite|improve this answer























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










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










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.






share|cite|improve this answer





















  • Thank you! Means a lot, enjoy the rest of your day.
    – Bill
    Jul 23 at 18:15














up vote
2
down vote



accepted










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.






share|cite|improve this answer





















  • Thank you! Means a lot, enjoy the rest of your day.
    – Bill
    Jul 23 at 18:15












up vote
2
down vote



accepted







up vote
2
down vote



accepted






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.






share|cite|improve this answer













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.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 23 at 18:06









J.G.

13.2k11424




13.2k11424











  • Thank you! Means a lot, enjoy the rest of your day.
    – Bill
    Jul 23 at 18:15
















  • Thank you! Means a lot, enjoy the rest of your day.
    – Bill
    Jul 23 at 18:15















Thank you! Means a lot, enjoy the rest of your day.
– Bill
Jul 23 at 18:15




Thank you! Means a lot, enjoy the rest of your day.
– Bill
Jul 23 at 18:15










up vote
2
down vote













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



$$frac43,frac4a3, frac4a^23,frac4a^33$$



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.






share|cite|improve this answer























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














up vote
2
down vote













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



$$frac43,frac4a3, frac4a^23,frac4a^33$$



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.






share|cite|improve this answer























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












up vote
2
down vote










up vote
2
down vote









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



$$frac43,frac4a3, frac4a^23,frac4a^33$$



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.






share|cite|improve this answer















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



$$frac43,frac4a3, frac4a^23,frac4a^33$$



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.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 23 at 18:04


























answered Jul 23 at 17:59









mrtaurho

750219




750219











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
















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















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




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












 

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