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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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
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@dan815 @Michele_Laino @perl @jigglypuff314 QH question for you guys :))
and until they come, do you know what a geometric sequence is?
For a geometric series we are multiplying by a number oover and over

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