Question

Find the sum of the first $20$ terms of the geometric series $\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+.....$

Answer 1

To find the sum of first $20$ terms of the geometric series $\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+........$, we have to first find the common ratio $r$.

In the given geometric series, the first term is $a_1=\frac{5}{2}$ and the second term is $a_2=\frac{5}{6}$ and so on.

We find the common ratio $r$ by dividing the second term by first term as shown below:

$r=\frac{\frac{5}{6}}{\frac{5}{2}}=\frac{5}{6}\times \frac{2}{5}=\frac{1}{3}<1$

We know that the sum of an geometric series with first term $a$ and common ratio $r$ is $S_n=\frac{a(1-r^n)}{1-r}$ if $r<1$

Now, to find the sum of first $20$ terms, substitute $a=\frac{5}{2},r=\frac{1}{3}$ and $n=20$ in $S_n=\frac{a(1-r^n)}{1-r}$ as follows:

$S_{20}=\frac{\frac{5}{2}[1-{\left(\frac{1}{3}\right)}^{20}]}{1-\frac{1}{3}}=\frac{\frac{5}{2}[1-{\left(\frac{1}{3}\right)}^{20}]}{\frac{3-1}{3}}=\frac{\frac{5}{2}[1-{\left(\frac{1}{3}\right)}^{20}]}{\frac{2}{3}}=\frac{5}{2}\times \frac{3}{2}[1-{\left(\frac{1}{3}\right)}^{20}]=\frac{15}{4}[1-{\left(\frac{1}{3}\right)}^{20}]$

Hence the sum is $\frac{15}{4}[1-{\left(\frac{1}{3}\right)}^{20}]$.