# How do you write the nth term rule for the sequence 6,14,22,30,38,...?

Nov 5, 2016

${a}_{n} = 8 n - 2$

#### Explanation:

There are 2 approaches. The second one is more accepted approach

$\textcolor{b l u e}{\text{First approach - should not be presented as a solution}}$

Observe that we have:

$n \text{ sequence build}$

$1 \text{ } \to 6 \textcolor{w h i t e}{1} = 6 + 0$
$2 \text{ } \to 14 = 6 + 8$
$3 \text{ } \to 22 = 6 + 8 + 8$
$4 \text{ } \to 30 = 6 + 8 + 8 + 8$
$5 \text{ } \to 38 = 6 + 8 + 8 + 8 + 8$

Now compare the count of 8's to the values of $n$
Notice that we have $n - 1$ of them

So for any ${i}^{\text{th}}$ term we have the term ${a}_{i}$ where

${a}_{i} = 6 + 8 \left(i - 1\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Second approach - this is the one to present as a solution}}$

Note that $6$ is $8 - 2$ so substitute for all the sixes giving:

$n \text{ sequence build}$

$1 \text{ } \to 6 \textcolor{w h i t e}{1} = 8 - 2$
$2 \text{ } \to 14 = 8 + 8 - 2$
$3 \text{ } \to 22 = 8 + 8 + 8 - 2$
$4 \text{ } \to 30 = 8 + 8 + 8 + 8 - 2$
$5 \text{ } \to 38 = 8 + 8 + 8 + 8 + 8 - 2$

The question asks for the ${n}^{\text{th}}$ term

So for any ${n}^{\text{th}}$ term we have the term ${a}_{n}$ where

" "color(blue)(ul(bar( |color(white)(2/2)a_n=(ixx8)-2" "->" "a_n=8n-2color(white)(2/2)|))