Question

How do you find the sum of the infinite geometric series $27+9+3+1+.....$ ?

Answer 1

These types of problems are very easy to solve and can easily be done by just using the formula for infinite G.P series. We need to have a fair idea of progression and series. The general form for an infinite geometric progression series is $a+ar+a{{r}^{2}}+a{{r}^{3}}+.....$ and for such series the sum of the series is given according to the formula $\dfrac{a}{1-r}$ . Here ‘a’ is the first term of the geometric progression series and ‘r’ is known as the common ratio. The common ratio of any G.P is defined as the ratio of the second and first term.

Complete step by step answer:
Now we start off with our solution. First of all we take $27$ common, and write the rest of the terms by assuming the sum value to $S$,
$S=27\left( 1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+..... \right)$ . Now, we consider $P=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+.....$ as our geometric series and it is related to $S$ as $S=27P$ .
Using the formulae of summation, we can easily find out the value of $P$ as,
$\dfrac{a}{1-r}$ , where $a=1$ and $r=\dfrac{1}{3}$ , Thus the value of $P$ is,

& P=\dfrac{1}{1-\dfrac{1}{3}} \\\ & \Rightarrow P=\dfrac{1}{\dfrac{2}{3}} \\\ & \Rightarrow P=\dfrac{3}{2} \\\ \end{aligned}$$ Now we can easily find the value of $$S$$ . We write, $$\begin{aligned} & S=27P \\\ & \Rightarrow S=27\times \dfrac{3}{2} \\\ & \Rightarrow S=\dfrac{81}{2} \\\ & \Rightarrow S=40.5 \\\ \end{aligned}$$ Thus, our answer to the problem is $$40.5$$. **Note:** One of the most common mistake that we need to keep in mind while solving infinite G.P problems is that, the total summation formula of the G.P that is $$\dfrac{a}{1-r}$$ , can only be applied when $$\left| r \right|<1$$ . For any other values of $$r$$ , the formula is not defined, as the series won’t even converge. We should also very carefully find the common ratio for the problem, failing which might lead to error in the solution.