How do you write a rule for the nth term of the arithmetic sequence given $a_6=13, a_14=25$ ?

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Answer 1 · 2018-04-19T15:38:07

$color(blue)(4+3/2n)$

The nth term of an arithmetic sequence is given as:

$a+(n-1)d$

Where $bba$ is the first term, $bbd$ is the common difference and $bbn$ is the nth term.

We need to find the first term and the common difference:

$a_6=13=>a+(6-1)d=13$

$a_14=25=>a+(14-1)d=25$

$a+5d=13 \ \ \ [1]$

$a+13d=25 \ \ \ \ [2]$

Solving simultaneously:

Subtract $[1]$ form $[2]$

$0+8d=12=>d=3/2$

Substituting this in $[1]$

$a+5(3/2)=13$

$a=13-3(3/2)=11/2$

For nth term:

$11/2+(n-1)(3/2)=11/2+3/2n-3/2=color(blue)(4+3/2n)$

How do you write a rule for the nth term of the arithmetic sequence given a_6=13, a_14=25a_6=13, a_14=25?