anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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TheSmartOne
  • TheSmartOne
@dan815 @Michele_Laino @perl @jigglypuff314 QH question for you guys :))
TheSmartOne
  • TheSmartOne
and until they come, do you know what a geometric sequence is?
dan815
  • dan815
For a geometric series we are multiplying by a number oover and over

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dan815
  • dan815
For arithmetic series there is a constant difference And for Geometric series there is a constant proportion
Michele_Laino
  • Michele_Laino
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
dan815
  • dan815
|dw:1435079880444:dw|
Michele_Laino
  • Michele_Laino
substituting your data we can write: \[\Large \left\{ \begin{gathered} 3 = {a_1}{q^3} \hfill \\ - 6 = {a_1}{q^9} \hfill \\ \end{gathered} \right.\]
anonymous
  • anonymous
Still waiting patiently for an answer
dan815
  • dan815
: )
TheSmartOne
  • TheSmartOne
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 )
anonymous
  • anonymous
anybody plz
Michele_Laino
  • Michele_Laino
dividing side by side those equations, we get: \[\Large {q^6} = - 2\]
Michele_Laino
  • Michele_Laino
please can you check your data, since the last equation has not solutions for q
anonymous
  • anonymous
what do you mean?
Michele_Laino
  • Michele_Laino
maybe a_10=6?
anonymous
  • anonymous
it's -6
Michele_Laino
  • Michele_Laino
I try to ask to another helper, please wait
Michele_Laino
  • Michele_Laino
@Australopithecus please help
Michele_Laino
  • Michele_Laino
@ganeshie8 please help
Michele_Laino
  • Michele_Laino
@dan815 please help
Michele_Laino
  • Michele_Laino
@chmvijay please help
anonymous
  • anonymous
what # raised to the power of 6 that gives 2
Michele_Laino
  • Michele_Laino
these equations are right: \[\Large \left\{ \begin{gathered} 3 = {a_1}{q^3} \hfill \\ - 6 = {a_1}{q^9} \hfill \\ \end{gathered} \right.\]
anonymous
  • anonymous
yes but i need the equation for the 22nd term. q is just the ratio
Michele_Laino
  • Michele_Laino
we have: \[\Large {a_{22}} = {a_1}{q^{21}}\]
anonymous
  • anonymous
if it was that easy, i would have solved already. Sorry
Michele_Laino
  • Michele_Laino
problems on geometrical sequences are always easy, I think that there is a typo into your textbook
anonymous
  • anonymous
there is no typo
Michele_Laino
  • Michele_Laino
as you can see we are not able to determine your sequence
anonymous
  • anonymous
when the teacher gives us the answer, i'll let you know
Michele_Laino
  • Michele_Laino
ok! I will wait that answer, thanks!
Michele_Laino
  • Michele_Laino
@mathmath333 please help
Michele_Laino
  • Michele_Laino
@misty1212 please help
mathmath333
  • mathmath333
the underlined word seems a mistake. as far my experience. |dw:1435082182285:dw|
mathmath333
  • mathmath333
in case if the OP is sure about the question then it should be \(\large \color{black}{\begin{align} &ar^{20}\hspace{.33em}\\~\\ &a \rightarrow \text{first term } \hspace{.33em}\\~\\ &r \rightarrow \text{common ratio } \hspace{.33em}\\~\\ \end{align}}\)
TheSmartOne
  • TheSmartOne
Do you mean \(\sf ar^{21}\) ?
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} &ar^{21}\hspace{.33em}\\~\\ &a \rightarrow \text{first term } \hspace{.33em}\\~\\ &r \rightarrow \text{common ratio } \hspace{.33em}\\~\\ \end{align}}\) this one.
TheSmartOne
  • TheSmartOne
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.
hartnn
  • hartnn
can you post the screenshot of your question ?
Australopithecus
  • Australopithecus
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.
Australopithecus
  • Australopithecus
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 \[a_{22} = a_{10}(r^{6})^2\] \[you\ know\ what\ a_{10} is\ so\ you\ should\ be\ able\ t o\ solve\ a_{22}\ or\ the\ 22nd\ term \]
Australopithecus
  • Australopithecus
This should give you the answer you seek
Australopithecus
  • Australopithecus
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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