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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.
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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TheSmartOne
 one year ago
Best ResponseYou've already chosen the best response.4@dan815 @Michele_Laino @perl @jigglypuff314 QH question for you guys :))

TheSmartOne
 one year ago
Best ResponseYou've already chosen the best response.4and until they come, do you know what a geometric sequence is?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2For a geometric series we are multiplying by a number oover and over

dan815
 one year ago
Best ResponseYou've already chosen the best response.2For arithmetic series there is a constant difference And for Geometric series there is a constant proportion

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2substituting 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
 one year ago
Best ResponseYou've already chosen the best response.0Still waiting patiently for an answer

TheSmartOne
 one year ago
Best ResponseYou've already chosen the best response.4Hi, @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 )

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2dividing side by side those equations, we get: \[\Large {q^6} =  2\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2please can you check your data, since the last equation has not solutions for q

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I try to ask to another helper, please wait

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2@Australopithecus please help

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2@ganeshie8 please help

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2@dan815 please help

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2@chmvijay please help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what # raised to the power of 6 that gives 2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2these equations are right: \[\Large \left\{ \begin{gathered} 3 = {a_1}{q^3} \hfill \\  6 = {a_1}{q^9} \hfill \\ \end{gathered} \right.\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes but i need the equation for the 22nd term. q is just the ratio

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we have: \[\Large {a_{22}} = {a_1}{q^{21}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if it was that easy, i would have solved already. Sorry

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2problems on geometrical sequences are always easy, I think that there is a typo into your textbook

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2as you can see we are not able to determine your sequence

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when the teacher gives us the answer, i'll let you know

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2ok! I will wait that answer, thanks!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2@mathmath333 please help

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2@misty1212 please help

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0the underlined word seems a mistake. as far my experience. dw:1435082182285:dw

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0in 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
 one year ago
Best ResponseYou've already chosen the best response.4Do you mean \(\sf ar^{21}\) ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} &ar^{21}\hspace{.33em}\\~\\ &a \rightarrow \text{first term } \hspace{.33em}\\~\\ &r \rightarrow \text{common ratio } \hspace{.33em}\\~\\ \end{align}}\) this one.

TheSmartOne
 one year ago
Best ResponseYou've already chosen the best response.4and 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/geometricsequencecalculator.html Scroll down to where it says Example Problems.

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0can you post the screenshot of your question ?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0This 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
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.0This should give you the answer you seek

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0If 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: \[a_n = a_1r^{n1}\] we get: \[a_{10} = a_1 r^{101} = 6\] \[a_{4} = a_1 r^{41} = 3\] 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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