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

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- jamiebookeater

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- TheSmartOne

@dan815 @Michele_Laino @perl @jigglypuff314 QH question for you guys :))

- TheSmartOne

and until they come, do you know what a geometric sequence is?

- dan815

For a geometric series we are multiplying by a number oover and over

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## More answers

- dan815

For arithmetic series there is a constant difference
And for Geometric series there is a constant proportion

- 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

|dw:1435079880444:dw|

- 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

Still waiting patiently for an answer

- dan815

: )

- 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

anybody plz

- Michele_Laino

dividing side by side those equations, we get:
\[\Large {q^6} = - 2\]

- Michele_Laino

please can you check your data, since the last equation has not solutions for q

- anonymous

what do you mean?

- Michele_Laino

maybe a_10=6?

- anonymous

it's -6

- Michele_Laino

I try to ask to another helper, please wait

- Michele_Laino

@Australopithecus please help

- Michele_Laino

@ganeshie8 please help

- Michele_Laino

@dan815 please help

- Michele_Laino

@chmvijay please help

- anonymous

what # raised to the power of 6 that gives 2

- 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

yes but i need the equation for the 22nd term. q is just the ratio

- Michele_Laino

we have:
\[\Large {a_{22}} = {a_1}{q^{21}}\]

- anonymous

if it was that easy, i would have solved already. Sorry

- Michele_Laino

problems on geometrical sequences are always easy, I think that there is a typo into your textbook

- anonymous

there is no typo

- Michele_Laino

as you can see we are not able to determine your sequence

- anonymous

when the teacher gives us the answer, i'll let you know

- Michele_Laino

ok! I will wait that answer, thanks!

- Michele_Laino

@mathmath333 please help

- Michele_Laino

@misty1212 please help

- mathmath333

the underlined word seems a mistake. as far my experience.
|dw:1435082182285:dw|

- 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

Do you mean \(\sf ar^{21}\) ?

- 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

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

can you post the screenshot of your question ?

- 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

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

This should give you the answer you seek

- 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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