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

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Answer 1 · 2016-11-05T11:01:05

$a_n=8n-2$

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

$color(blue)("First approach - should not be presented as a solution")$

Observe that we have:

$n" sequence build"$

$1" "->6color(white)(1) = 6+0$
$2" "->14 = 6+8$
$3" "->22=6+8+8$
$4" "->30=6+8+8+8$
$5" "->38=6+8+8+8+8$

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

So for any $i^("th")$ term we have the term $a_i$ where

$a_i=6+8(i-1)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Second approach - this is the one to present as a solution")$

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

$n" sequence build"$

$1" "->6color(white)(1) = 8-2$
$2" "->14 = 8+8-2$
$3" "->22=8+8+8-2$
$4" "->30=8+8+8+8-2$
$5" "->38=8+8+8+8+8-2$

The question asks for the $n^("th")$ term

So for any $n^("th")$ term we have the term $a_n$ where

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

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