use the formula to evaluate the series
3-9+27-81+...-a8
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jimthompson5910 (jim_thompson5910):
post the screenshot please
OpenStudy (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?
OpenStudy (anonymous):
i dont know how to do screen shot
OpenStudy (anonymous):
jimthompson5910 (jim_thompson5910):
thanks
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OpenStudy (anonymous):
okay now i know
jimthompson5910 (jim_thompson5910):
each term is found by multiplying the previous term by -3
jimthompson5910 (jim_thompson5910):
3*(-3) = -9
-9*(-3) = 27
27*(-3) = -81
etc etc
OpenStudy (anonymous):
yeah
jimthompson5910 (jim_thompson5910):
generate the first 8 terms, then add them up
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OpenStudy (anonymous):
okay hold on lets see
OpenStudy (anonymous):
-4920
jimthompson5910 (jim_thompson5910):
alternatively, you can use this formula
\[\Large S_n = a*\frac{1-r^n}{1-r}\]
OpenStudy (anonymous):
yeah thats what its says to use
jimthompson5910 (jim_thompson5910):
yeah it's -4920
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OpenStudy (anonymous):
okay lets see ill send you a screen shot if its wrong okay
OpenStudy (anonymous):
you are amazing
jimthompson5910 (jim_thompson5910):
ok I'm glad that one worked
OpenStudy (anonymous):
okay i have another
OpenStudy (anonymous):
i have a question are you a teacher for algebra
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OpenStudy (anonymous):
because you seem to know what your doing
jimthompson5910 (jim_thompson5910):
yes I've had lots and lots of practice
OpenStudy (anonymous):
sweet heres the other question
OpenStudy (anonymous):
jimthompson5910 (jim_thompson5910):
this is a geometric series
what's the first term? common ratio?
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OpenStudy (anonymous):
1/3
jimthompson5910 (jim_thompson5910):
first term = 1/3, yes
OpenStudy (anonymous):
its going up in numerator x2 denomanator x3
jimthompson5910 (jim_thompson5910):
common ratio = -2/3
OpenStudy (anonymous):
okay
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OpenStudy (anonymous):
so its diverge
OpenStudy (anonymous):
right
jimthompson5910 (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
jimthompson5910 (jim_thompson5910):
Rule:
if |r| < 1, then the series converges
otherwise, the series diverges
OpenStudy (anonymous):
okay so its fixed sum is 1/5
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jimthompson5910 (jim_thompson5910):
if you generated all of the terms and added them up, they would add up to 1/5
OpenStudy (anonymous):
okay
OpenStudy (anonymous):
so converges are out right
jimthompson5910 (jim_thompson5910):
what do you mean? I wrote above that the series converges