Answer:
Arithmetic with common difference of [tex]\sf \frac{1}{3}[/tex]
Step-by-step explanation:
[tex]\textsf{Given sequence}=4, \dfrac{13}{3}, \dfrac{14}{3}, 5, \dfrac{16}{3},...[/tex]
If a sequence is arithmetic, the difference between consecutive terms is the same (this is called the common difference).
If a sequence is geometric, the ratio between consecutive terms is the same (this is called the common ratio).
[tex]\sf 4\quad \overset{+\frac{1}{3}}{\longrightarrow}\quad\dfrac{13}{3}\quad \overset{+\frac{1}{3}}{\longrightarrow}\quad \dfrac{14}{3}\quad \overset{+\frac{1}{3}}{\longrightarrow}\quad 5\quad \overset{+\frac{1}{3}}{\longrightarrow}\quad \dfrac{16}{3}[/tex]
As the difference between consecutive terms is [tex]\sf \frac{1}{3}[/tex] then the sequence is arithmetic with common difference of [tex]\sf \frac{1}{3}[/tex]
General form of an arithmetic sequence: [tex]\sf a_n=a+(n-1)d[/tex]
where:
- [tex]\sf a_n[/tex] is the nth term
- a is the first term
- d is the common difference between terms
Given:
- a = 4
- [tex]\sf d=\dfrac{1}{3}[/tex]
So the formula for the nth term of this sequence is:
[tex]\implies \sf a_n=4+(n-1)\dfrac{1}{3}[/tex]
[tex]\implies \sf a_n=\dfrac{1}{3}n+\dfrac{11}{3}[/tex]
