# Question #c18f2

Apr 13, 2017

$7$ terms in total:
$27 + 9 + 3 + 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} = \frac{1093}{27}$

#### Explanation:

The series is a geometric series and is thus in the form ${a}_{n} = {a}_{n - 1} \cdot r$ or ${a}_{n} = {a}_{1} \cdot {r}^{n - 1}$.

The first term, ${a}_{1}$, is $27$. The common ratio $r$ is $\frac{1}{3}$.

We need to find $n$ such that ${\sum}_{k = 1}^{n} 27 \cdot {\left(\frac{1}{3}\right)}^{k - 1} = \frac{1093}{27}$.

The sum of a geometric series is equal to $\frac{{a}_{1} \left(1 - {r}^{n - 1}\right)}{1 - r}$. Thus, the left-hand side can be expressed as $\frac{27 \left(1 - {\left(\frac{1}{3}\right)}^{n - 1}\right)}{1 - \frac{1}{3}} = \frac{81}{2} \left(1 - \frac{1}{3} ^ \left(n - 1\right)\right)$.

Solve $\frac{81}{2} \left(1 - \frac{1}{3} ^ \left(n - 1\right)\right) = \frac{1093}{27}$. This becomes $\frac{1}{3} ^ \left(n - 1\right) = 1 - \frac{2186}{2187} = \frac{1}{2187}$. Solving for $n$ results in $n = 7$.