use the formula to evaluate the series
3-9+27-81+...-a8

- anonymous

use the formula to evaluate the series
3-9+27-81+...-a8

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- schrodinger

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- jim_thompson5910

post the screenshot please

- triciaal

there are 2 types of series arithmetic which has a common difference from one term to the next and the geometric progression that has a common ratio from on term to the next.
can you now identify what kind of series this is?

- anonymous

i dont know how to do screen shot

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## More answers

- anonymous

##### 1 Attachment

- jim_thompson5910

thanks

- anonymous

okay now i know

- jim_thompson5910

each term is found by multiplying the previous term by -3

- jim_thompson5910

3*(-3) = -9
-9*(-3) = 27
27*(-3) = -81
etc etc

- anonymous

yeah

- jim_thompson5910

generate the first 8 terms, then add them up

- anonymous

okay hold on lets see

- anonymous

-4920

- jim_thompson5910

alternatively, you can use this formula
\[\Large S_n = a*\frac{1-r^n}{1-r}\]

- anonymous

yeah thats what its says to use

- jim_thompson5910

yeah it's -4920

- anonymous

okay lets see ill send you a screen shot if its wrong okay

- anonymous

you are amazing

- jim_thompson5910

ok I'm glad that one worked

- anonymous

okay i have another

- anonymous

i have a question are you a teacher for algebra

- anonymous

because you seem to know what your doing

- jim_thompson5910

yes I've had lots and lots of practice

- anonymous

sweet heres the other question

- anonymous

##### 1 Attachment

- jim_thompson5910

this is a geometric series
what's the first term? common ratio?

- anonymous

1/3

- jim_thompson5910

first term = 1/3, yes

- anonymous

its going up in numerator x2 denomanator x3

- jim_thompson5910

common ratio = -2/3

- anonymous

okay

- anonymous

so its diverge

- anonymous

right

- jim_thompson5910

r = -2/3
since |r| < 1 is true, this means the infinite series does converge and there is a fixed sum it reaches
that sum would be
S = a/(1-r)
S = (1/3)/(1-(-2/3))
S = (1/3)/(5/3)
S = (1/3)*(3/5)
S = 1/5

- jim_thompson5910

Rule:
if |r| < 1, then the series converges
otherwise, the series diverges

- anonymous

okay so its fixed sum is 1/5

- jim_thompson5910

if you generated all of the terms and added them up, they would add up to 1/5

- anonymous

okay

- anonymous

so converges are out right

- jim_thompson5910

what do you mean? I wrote above that the series converges

- jim_thompson5910

since |r|<1

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