Answer
See below
Work Step by Step
Given: $d_n=\frac{1}{2}n(n-3)$
The first six terms of the sequence:
$a_3=\frac{1}{2}\times3\times(3-3)=0\\a_4=\frac{1}{2}\times4\times(4-3)=2\\a_5=\frac{1}{2}\times5\times(5-3)=5\\a_6=\frac{1}{2}\times6\times(6-3)=9\\a_7=\frac{1}{2}\times7\times(7-3)=14\\a_8=\frac{1}{2}\times8\times(8-3)=20$
Determine the differences between the terms:
$$a_4-a_3=2-0=2\\a_5-a_4=5-2=3$$
Since they are not constant, the sequence is not arithmetic.
Find the quotients: $\frac{a_4}{a_3}=\frac{2}{0}=\emptyset\\\frac{a_5}{a_4}=\frac{5}{2}=2.5\\\frac{a_6}{a_5}=\frac{9}{5}=1.8$
Hence, the sequence is neither geometric nor arithmetic.
