Answer
Neither arithmetic, nor geometric
Work Step by Step
We are given the sequence:
$\left\{\dfrac{2n-3}{2n+1}\right\}$
Determine the difference of consecutive terms:
$a_2-a_1=\dfrac{2(2)-3}{2(2)+3}-\dfrac{2(1)-3}{2(1)+3}=\dfrac{1}{7}-\left(-\dfrac{1}{5}\right)=\dfrac{12}{35}$
$a_3-a_2=\dfrac{2(3)-3}{2(3)+3}-\dfrac{2(2)-3}{2(2)+3}=\dfrac{3}{9}-\dfrac{1}{7}=\dfrac{1}{3}-\dfrac{1}{7}=\dfrac{4}{21}$
As the difference is not constant, the sequence is not arithmetic.
Determine the ratio of consecutive terms:
$\dfrac{a_2}{a_1}=\dfrac{\dfrac{2(2)-3}{2(2)+3}}{\dfrac{2(1)-3}{2(1)+3}}=\dfrac{\dfrac{1}{7}}{-\dfrac{1}{5}}=-\dfrac{5}{7}$
$\dfrac{a_3}{a_2}=\dfrac{\dfrac{2(3)-3}{2(3)+3}}{\dfrac{2(2)-3}{2(2)+3}}=\dfrac{\dfrac{3}{9}}{\dfrac{1}{7}}=\dfrac{7}{3}$
As the ratio of consecutive terms is not constant, the sequence is not geometric.
Therefore the sequence is neither arithmetic, nor geometric.
