Given:-
[tex]\frac{5}{6},\frac{2}{3},\frac{1}{2},\frac{1}{3},\frac{1}{6}[/tex]
To find the given sequence is arithmetic or geometric or neither.
If the given sequence is arithmetic, then it obeys the formula,
[tex]t_2-t_1=t_3-t_2=t_4-t_3[/tex]
So now we substitute the values. So we get,
[tex]\begin{gathered} \frac{2}{3}-\frac{5}{6}=\frac{1}{2}-\frac{2}{3} \\ \frac{4}{6}-\frac{5}{6}=\frac{3-4}{6} \\ -\frac{1}{6}-\frac{1}{6} \end{gathered}[/tex]
Now we do this for the next two terms,
[tex]\begin{gathered} \frac{1}{2}-\frac{2}{3}=\frac{1}{3}-\frac{1}{2} \\ \frac{3-4}{6}=\frac{2-3}{6} \\ -\frac{1}{6}=-\frac{1}{6} \end{gathered}[/tex]
Now we do this for the next two terms,
[tex]\begin{gathered} \frac{1}{3}-\frac{1}{2}=\frac{1}{6}-\frac{1}{3} \\ \frac{2-3}{6}=\frac{1-2}{6} \\ -\frac{1}{6}=-\frac{1}{6} \end{gathered}[/tex]
So the given sequence is ARITHMETIC SEQUENCE.
So the common difference is,
[tex]d=-\frac{1}{6}[/tex]
