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anonymous

  • one year ago

What is the equation for the 22nd term of the geometric sequence in which the fourth term is 3 and the 10th term is -6.

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  1. TheSmartOne
    • one year ago
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    @dan815 @Michele_Laino @perl @jigglypuff314 QH question for you guys :))

  2. TheSmartOne
    • one year ago
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    and until they come, do you know what a geometric sequence is?

  3. dan815
    • one year ago
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    For a geometric series we are multiplying by a number oover and over

  4. dan815
    • one year ago
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    For arithmetic series there is a constant difference And for Geometric series there is a constant proportion

  5. Michele_Laino
    • one year ago
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    we have the subsequent formulas: \[\Large \begin{gathered} {a_4} = {a_1}{q^3} \hfill \\ {a_{10}} = {a_1}{q^9} \hfill \\ \end{gathered} \] where a_1, a_4, a_10 are the first, fourth and tenth terms, furthermore q is the cinstant of our sequence

  6. dan815
    • one year ago
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    |dw:1435079880444:dw|

  7. Michele_Laino
    • one year ago
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    substituting your data we can write: \[\Large \left\{ \begin{gathered} 3 = {a_1}{q^3} \hfill \\ - 6 = {a_1}{q^9} \hfill \\ \end{gathered} \right.\]

  8. anonymous
    • one year ago
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    Still waiting patiently for an answer

  9. dan815
    • one year ago
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    : )

  10. TheSmartOne
    • one year ago
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    Hi, @NegLakay Welcome to OpenStudy! OpenStudy is here for you to learn, not give you answers. So patiently answer the hints that the QH's (Qualified Helpers) give you and they will direct you to your answer step by step so you can learn how to solve it )

  11. anonymous
    • one year ago
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    anybody plz

  12. Michele_Laino
    • one year ago
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    dividing side by side those equations, we get: \[\Large {q^6} = - 2\]

  13. Michele_Laino
    • one year ago
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    please can you check your data, since the last equation has not solutions for q

  14. anonymous
    • one year ago
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    what do you mean?

  15. Michele_Laino
    • one year ago
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    maybe a_10=6?

  16. anonymous
    • one year ago
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    it's -6

  17. Michele_Laino
    • one year ago
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    I try to ask to another helper, please wait

  18. Michele_Laino
    • one year ago
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    @Australopithecus please help

  19. Michele_Laino
    • one year ago
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    @ganeshie8 please help

  20. Michele_Laino
    • one year ago
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    @dan815 please help

  21. Michele_Laino
    • one year ago
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    @chmvijay please help

  22. anonymous
    • one year ago
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    what # raised to the power of 6 that gives 2

  23. Michele_Laino
    • one year ago
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    these equations are right: \[\Large \left\{ \begin{gathered} 3 = {a_1}{q^3} \hfill \\ - 6 = {a_1}{q^9} \hfill \\ \end{gathered} \right.\]

  24. anonymous
    • one year ago
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    yes but i need the equation for the 22nd term. q is just the ratio

  25. Michele_Laino
    • one year ago
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    we have: \[\Large {a_{22}} = {a_1}{q^{21}}\]

  26. anonymous
    • one year ago
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    if it was that easy, i would have solved already. Sorry

  27. Michele_Laino
    • one year ago
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    problems on geometrical sequences are always easy, I think that there is a typo into your textbook

  28. anonymous
    • one year ago
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    there is no typo

  29. Michele_Laino
    • one year ago
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    as you can see we are not able to determine your sequence

  30. anonymous
    • one year ago
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    when the teacher gives us the answer, i'll let you know

  31. Michele_Laino
    • one year ago
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    ok! I will wait that answer, thanks!

  32. Michele_Laino
    • one year ago
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    @mathmath333 please help

  33. Michele_Laino
    • one year ago
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    @misty1212 please help

  34. mathmath333
    • one year ago
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    the underlined word seems a mistake. as far my experience. |dw:1435082182285:dw|

  35. mathmath333
    • one year ago
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    in case if the OP is sure about the question then it should be \(\large \color{black}{\begin{align} &ar^{20}\hspace{.33em}\\~\\ &a \rightarrow \text{first term } \hspace{.33em}\\~\\ &r \rightarrow \text{common ratio } \hspace{.33em}\\~\\ \end{align}}\)

  36. TheSmartOne
    • one year ago
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    Do you mean \(\sf ar^{21}\) ?

  37. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} &ar^{21}\hspace{.33em}\\~\\ &a \rightarrow \text{first term } \hspace{.33em}\\~\\ &r \rightarrow \text{common ratio } \hspace{.33em}\\~\\ \end{align}}\) this one.

  38. TheSmartOne
    • one year ago
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    and so far, I've done the problem. But I've arrived at the same solution as Michele unfortunately. And to make sure we were doing it correct, I even looked for an example and it shows the same steps we took: http://calculator.tutorvista.com/geometric-sequence-calculator.html Scroll down to where it says Example Problems.

  39. hartnn
    • one year ago
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    can you post the screenshot of your question ?

  40. Australopithecus
    • one year ago
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    This cant be solved without complex numbers because either for all even n points they would have to be all positive or all negative. In your case you have two even n values with different signs this is not possible unless the sequence contains complex numbers.

  41. Australopithecus
    • one year ago
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    You know already, \[r = -(-2)^{\frac{1}{6}}\] or \[r = (-2)^{\frac{1}{6}}\] you can use either and note: \[a_{22} = a_{10}r^{12}\] or \[a_{22} = a_{10}(r^{6})^2\] \[you\ know\ what\ a_{10} is\ so\ you\ should\ be\ able\ t o\ solve\ a_{22}\ or\ the\ 22nd\ term \]

  42. Australopithecus
    • one year ago
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    This should give you the answer you seek

  43. Australopithecus
    • one year ago
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    If you are confused about how r was acquired it is just the ratio of these two terms (you can check the tutor site or look at @Michele_Laino answers to this question he explains it as well but he uses q instead of r (just saying q = r) Anyways Using the formula: \[a_n = a_1r^{n-1}\] we get: \[a_{10} = a_1 r^{10-1} = -6\] \[a_{4} = a_1 r^{4-1} = 3\] then we take the ratio of these two terms to figure out what r is \[\frac{a_{10}}{a_4} =\frac{a_1r^{9}}{a_1r^{9}} = \frac{-6}{3}\] notice a_1 cancels out and doing simple algebra you can solve r which will be (-2)^(1/6) or -(-2)^(1/6) in this solution you can use ether r value and you will end up with the same value for a_22 Hope this was helpful

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