Answer
First five terms: $8,10,12,14, 16$.
The sequence is arithmetic with common difference $2$.
Work Step by Step
Given: $a_n=2(n+3)$
To find the first five terms of the given sequence, substitute $n=1,2,3,4,5$ into the given formula to obtain: \begin{align*} a_1&=2(1+3)=8\\ \\a_2&=2(2+3) =10\\ \\a_3&=2(3+3)=12\\ \\a_4&=2(4+3)=14\\ \\a_5&=2(5+3)=16\end{align*}
In order to determine the sequence to be geometric, the quotient of all consecutive terms must be constant. Note that: \begin{align*} \dfrac{a_2}{a_1}&=\dfrac{10}{8}=\dfrac{5}{4}\\ \\\dfrac{a_3}{a_2}&=\dfrac{12}{10}= \dfrac{6}{5} \end{align*} The quotients are not the same so the sequence is not geometric.
In order to determine the sequence to be arithmetic, the difference of all consecutive terms must be constant. Note that \begin{align*} a_2-a_1&=10-8=2\\ \\a_3-a_2&=12-10=2\end{align*} There is a constant difference of $2$, so the sequence is arithmetic.
Therefore, the given sequence arithmetic sequence with common difference $2$.
