Answer:
option d) [tex]a(n)=4(n-1)+\frac{3}{4}[/tex]
step by step explanation:
Given sequence is [tex]a_n={4, \frac{19}{4},\frac{11}{2},\frac{25}{4},7,...}[/tex]
Let [tex]a_1=4[/tex], [tex]a_2=\frac{19}{4}[/tex], [tex]a_3=\frac{11}{2}[/tex], [tex]a_4=\frac{25}{4}[/tex],[tex]a_5=4[/tex], ...
common difference [tex]d=a_2-a_1[/tex]
[tex]d=\frac{19}{4}-4[/tex]
[tex]d=\frac{19-16}{4}[/tex]
[tex]d=\frac{3}{4}[/tex]
[tex]d=a_3-a_2[/tex]
[tex]d=\frac{11}{2}-\frac{19}{4}[/tex]
[tex]d=\frac{22-19}{4}[/tex]
[tex]d=\frac{3}{4}[/tex]
Therefore the common difference [tex]d=\frac{3}{4}[/tex]
The recurrsive formula for arithmetic sequence is
[tex]a(n)=a(n-1)+d[/tex]
Therefore
[tex]a(n)=4(n-1)+\frac{3}{4}[/tex] where a=4 and [tex]d=\frac{3}{4}[/tex]
