- anonymous

what are the explicit equation and domain for a geometric sequence with a first term of 5 and a second term of -10

- schrodinger

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- anonymous

- anonymous

- jim_thompson5910

5r = -10
r = ???

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## More answers

- anonymous

-2

- jim_thompson5910

First Term: a = 5
Common Ratio: r = -2
Plug those into
\[\Large a_{n} = a*(r)^{n-1}\]
to get the general nth term formula

- anonymous

5n = 5 x (-2) n-1
5n = -10 n-1
What do i do from here

- myininaya

no a_n symbolizes the n term of the sequence
a_n and a are different values
also we don't have (ar)^(n-1)
we just have that the r is to the (n-1)

- anonymous

Ok, so how do i set up this equation

- myininaya

\[a_n=a \cdot r^{n-1}\]
just replace the a with 5 and replace the r with -2 like jim said

- anonymous

5n = -10^n-1
?

- myininaya

again a_n and a are different values
a_n is a variable by itself
you can't multiply a and r
because r has an exponent

- anonymous

5_n = 5 x -2^n-1

- myininaya

you cannot replace a_n with 5_n
this makes no sense
let me show you how this formula is come up with so maybe you understand better
{a_n} is a sequence of numbers
those numbers are
a_1=5
a_2=5(-2)=-10
a_3=5(-2)(-2)=5(-2)^2=20
a_4=5(-2)(-2)(-2)=5(-2)^3=-40
a_5=5(-2)(-2)(-2)(-2)=5(-2)^4=80
...
therefore a_n=5(-2)(-2)(-2)(-2)(-2)(-2)...(-2)=
(by the way that is a (n-1) amount of (-2))
I put a (n-1) amount of (-2)
because looking at a_1 we have (-2)^0=(-2)^(1-1)
looking at a_2 we have (-2)^1=(-2)^(2-1)
looking at a_3 we have (-2)^2=(-2)^(3-1)
it is always one less than where the number is in the sequence

- anonymous

So how do i set up the equation ? im confused

- anonymous

Does the 5 not go where the n is

- jim_thompson5910

5 goes where 'a' is
but \(\Large a_n\) is NOT the same as just 'a'

- jim_thompson5910

If it confuses you, think of it as
\[\Large T = a*(r)^{n-1}\]
T = nth term
a = first term
r = common ratio
n = positive whole number (used to identify which term you're dealing with)
example: n = 3 ---> 3rd term

- anonymous

Ok

- jim_thompson5910

The notation \(\Large a_n\) is used for sequences because the 'n' changes to a positive whole number to indicate a certain term
example: 17th term ---> n = 17 ---> \(\Large a_n = a_{17}\)

- anonymous

Ok, so you leave the an as it is

- myininaya

Yep it is a formula for the nth term in the sequence

- anonymous

So it is an = 5 (-15)^n-1; all integers where n less than or equal to 1

- myininaya

how do you get -15 for r?

- myininaya

r symbolizes the geometric ratio
to find r all you have to do is evaluate a_2/a_1
or take any two consecutive numbers in the sequence and put the 2 nd term of those numbers over the 1st term of those numbers

- myininaya

also this was already determined way above when joe was working with you

- anonymous

the answer is A!!!!!!!!!!!!!!!!!

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