We have a series: 2 1/2, 3 3/4, ...
We can consider this is a arithmetic series, so it has a common difference. This difference can be calculated as:
[tex]\begin{gathered} d=a_2-a_1 \\ d=(3+\frac{3}{4})-(2+\frac{1}{2}) \\ d=3-2+\frac{3}{4}-\frac{1}{2} \\ d=1+\frac{3-2}{4} \\ d=1+\frac{1}{4} \end{gathered}[/tex]We then can consider that each term adds 1 1/4 to the previous term.
We can calculate the third and fourth term of the series as:
[tex]\begin{gathered} a_3=a_2+d \\ a_3=3+\frac{3}{4}+1+\frac{1}{4}=3+1+\frac{3+1}{4}=4+\frac{4}{4}=4+1=5 \end{gathered}[/tex][tex]\begin{gathered} a_4=a_3+d \\ a_4=5+1+\frac{1}{4}=6+\frac{1}{4} \end{gathered}[/tex]Answer: the third term is 5 and the fourth term is 6 1/4.
