Find the sum of the infinite geometric series: 3+2+frac{4}{3}+frac{8}{9}+...

Find the sum of the infinite geometric series: 3+2+frac{4}{3}+frac{8}{9}+...

Question
Series
asked 2021-02-09
Find the sum of the infinite geometric series:
\(3+2+\frac{4}{3}+\frac{8}{9}+...\)

Answers (1)

2021-02-10
Given: \(S=3+2+\frac{4}{3}+\frac{8}{9}+...\)
We know that sum(S) of infinite geometric series is given by
\(S=\frac{a}{1-r}\)
Where a is first term and r is common ratio
Here,
\(a=3,r=\frac{2}{3}\)
So, by using equation(1) sum(S) of infinite geometric series will be
\(S=\frac{3}{1-\frac{2}{3}}\)
\(S=\frac{3}{(\frac{3-2}{3})}\)
\(S=\frac{3}{(\frac{1}{3})}\)
\(=3(3)\)
\(=9\)
Hence, sum of given infinite geometric series is 9.
0

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