Question

Find the eleventh term of the sequence $7,14,28,56,112,224,\dots$Then find the general term for the sequence.

Short answer

The eleventh term of the sequence is$7168$ and the general term is equal to$a_n=7{\left(2\right)}^{n-1}$

Step-by-step solution

Step 1. Given information

The sequence is $7,14,28,56,112,224,...$

Step 2. Find the eleventh term of the sequence.

Consider the sequence:

$7,14,28,56,112,256,...$

Let $a,r$denote the first term and the common ratio respectively.

$\begin{array}{rcl}a& =& 7\\ r& =& \frac{14}{7}\\ & =& 2\end{array}$

$\begin{array}{rcl}{a}_n& =& a{r}^{n-1}\\ a_{11}& =& 7{\left(2\right)}^{11-1}\\ & =& 7{\left(2\right)}^{10}\\ & =& 7168\end{array}$

Step 3. Find the general term of the sequence.

Put $a=7,r=2$in$a_n=a{r}^{n-1}$.

$a_n=7{\left(2\right)}^{n-1}$