the one with 6 and 27 first then 5 and 50 then 6 and 33
Step-by-step explanation:
i took the test and got it right good luck plato family
In order from least to greatest based on their sums, the series can be written as -
Series 2 < Series 3 < Series 1
We have three series.
We have to arrange these series from least to greatest based on their sum.
The formula to find the sum of first ' n ' terms of a Geometric Progression is -
[tex]S_{n} =\frac{a(1-r^{n} )}{1-r} \\[/tex]
According to the question given, we have - Series 1, Series 2 and Series 3 -
[tex]\sum_{k=0}^{n}33 (\frac{1}{2}) ^{k} = 33 + \frac{33}{2} +\frac{33}{4} ....\\\sum_{k=0}^{n}27 (\frac{1}{3}) ^{k} = 27 + \frac{27}{3} +\frac{27}{9} ....\\\\\sum_{k=0}^{n}50 (\frac{1}{8}) ^{k} = 50 + \frac{50}{8} +\frac{50}{64} ....\\\\\\\sum_{k=0}^{6}33 (\frac{1}{2}) ^{k} = \frac{33(1- (0.5)^{6}) }{1 -0.5} =64.96\\\sum_{k=0}^{6}27 (\frac{1}{3}) ^{k} = \frac{27(1- (\frac{1}{3} )^{6}) }{1 -\frac{1}{3} } = 40.44\\\sum_{k=0}^{6}50 (\frac{1}{8}) ^{k} = \frac{33(1- (\frac{1}{8} )^{6}) }{1 -\frac{1}{8} } =57.14[/tex]
It can be seen from the above results that : Series 2 < Series 3 < Series 1
Hence, in order from least to greatest based on their sums, the series can be written as -
Series 2 < Series 3 < Series 1
To solve more questions on Geometric sequences, visit the link below -
https://brainly.com/question/26093201
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