Answer:
[tex]a_n=\frac{1}{2} (31-5n)=15.5-2.5n\\\\n\in N[/tex]
Step-by-step explanation:
As you can see the sequence starts in 13 and it decreases 2.5 units every time, so, we can assume that the sequence starts at 15.5:
[tex]a_1=15.5-2.5[/tex]
In this sense, we can infer that a possible sequence is:
[tex]a_n=15.5-2.5n=\frac{1}{2} (31-5n)\\\\n\in N[/tex]
Let's verify it:
[tex]a_1=15.5-2.5(1)=13\\a_2=15.5-2.5(2)=15.5-5=10.5\\a_3=15.5-2.5(3)=15.5-7.5=8\\a_4=15.5-2.5(4)=15.5-10=5.5\\a_5=15.5-2.5(5)=15.5-12.5=3\\a_6=15.5-2.5(6)=15.5-15=0.5[/tex]
