## Precalculus: Mathematics for Calculus, 7th Edition

Sequence is arithmetic $d=6$ $a_{n}=a+\left( n-1\right) d=-4+6\left( n-1\right)$
In order a sequence to be arithmetic $a_{n+1}-a_{n}=a_{n}-a_{n-1}\Rightarrow 2a_{n}=a_{n+1}+a_{n-1}$ $2a_{n}=2\left( 6n-10\right) =12n-20$ $a_{n+1}+a_{n-1}=6\left( n+1\right) -10+6\left( n-1\right) -10=6n+6-10+6n-6-10=12n-20$ $\Rightarrow 2a_{n}=a_{n+1}+a_{n-1}$ So this sequence is arithmetic $a_{n}=a+\left( n-1\right) d=a-d+nd=6n-10\Rightarrow d=6;$ $a-d=-10\Rightarrow a=-4$ So the sequence will be $a_{n}=a+\left( n-1\right) d=-4+6\left( n-1\right)$