Answer
First five terms: $\dfrac{1}{2}, \dfrac{2}{3}, \dfrac{3}{4}, \dfrac{4}{5}, \dfrac{5}{6}$
The sequence is neither geometric nor arithmetic.
Work Step by Step
Given: $a_n=\dfrac{n}{n+1}$
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&=\dfrac{1}{1+1}=\dfrac{1}{2}\\
\\a_2&=\dfrac{2}{2+1}=\dfrac{2}{3}\\
\\a_3&=\dfrac{3}{3+1}=\dfrac{3}{4}\\
\\a_4&=\dfrac{4}{4+1}=\dfrac{4}{5}\\
\\a_5&=\dfrac{5}{5+1}=\dfrac{5}{6}\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{\frac{2}{3}}{\frac{1}{2}}=\dfrac{2}{3}\cdot\dfrac{2}{1}=\dfrac{4}{3}\\
\\\dfrac{a_3}{a_2}&=\dfrac{\frac{3}{4}}{\frac{2}{3}}=\dfrac{3}{4} \cdot \dfrac{3}{2}=\dfrac{9}{8}
\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&=\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{4}{6}-\dfrac{3}{6}=\dfrac{1}{6}\\
\\a_3-a_2&=\dfrac{3}{4}-\dfrac{2}{3}=\dfrac{9}{12}-\dfrac{8}{12}=\dfrac{1}{12}
\end{align*}
There is no constant difference so the sequence is not arithmetic.
Therefore, the given sequence is neiher geometric nor arithmetic.
