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

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

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.

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