The population of a local species of dragonfly can be found using an infinite geometric series where a1 = 42 and the common ratio is three fourths. Write the sum in sigma notation and calculate the sum that will be the upper limit of this population.

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The population of a local species of dragonfly can be found using an infinite geometric series where a1 = 42 and the common ratio is three fourths. Write the sum in sigma notation and calculate the sum that will be the upper limit of this population.

Mathematics
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what do you have so far?
i have|dw:1442973187600:dw|

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Other answers:

you're very close
oh and the 42(3/4)^i-1?
it should be \[\Large \sum_{i=1}^{\infty} 42\left(\frac{3}{4}\right)^{i-1}\]
Yes! That's what I meant. Thank-you :)
now `calculate the sum that will be the upper limit of this population.`
Hmmm...it wouldn't be diveregent right?
divergent*
you see how a = 42 and r = 3/4 right?
yes
ok! So we are plugging it into the geometric series formula!
yeah since |r| < 1 is true, we can use S = a/(1-r)
ok. So I got 168?
me too
Can you help me on another?
sure
What is the sum of the arithmetic sequence 8, 14, 20 …, if there are 24 terms?
first term = ?? common difference = ??
A1=8 C.D.=-6
C.D is +6 actually
we add 6 to each term to get the next one
oh! Sorry :)
now use the formula \[\Large S_n = \frac{n(2a+d(n-1))}{2}\] n = number of terms added up a = first term d = common difference
So far, I have s24=24(16+6n-1/2)
n = 24, so replace the n in n-1 with 24
oh! Thanks
\[\Large S_n = \frac{n(2a+d(n-1))}{2}\] \[\Large S_{24} = \frac{24(2*8+6(24-1))}{2}\] \[\Large S_{24} = ???\]
Ok! Hold on lol
1848 :)
good

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