# For the sequence -3, -12, -48, … how do you find the sum of the first 12 terms?

Mar 7, 2016

Use the formula ${\sum}_{0}^{n} = \frac{a \left(1 - {r}^{n}\right)}{1 - r}$ where $a$ is the first term, $r$ is the common ratio and $n$ is the number of terms.

#### Explanation:

The sequence here is $- 3 , \left(- 3 \cdot 4\right) , \left(- 3 \cdot 4 \cdot 4\right) , \ldots \ldots$

So the $n$th term is $\left(- 3\right) \cdot {4}^{n - 1}$

The first term is $- 3$ and the common ratio is $4$.

${\sum}_{1}^{12} = \frac{\left(- 3\right) \left(1 - {4}^{12}\right)}{1 - 4} = \frac{\left(- 3\right) \left(- 16777215\right)}{- 3}$

$= - 16777215$