Question

Determine whether each of the following sequences are arithmetic, geometric or neither. If arithmetic, state the common difference. If geometric, state the common ratio.

4, 13/3, 14/3, 5, 16/3, …

Is the answer: Neither?

Answer 1

$\qquad\qquad\huge\underline{{\sf Answer}}$

$\textbf{Let's see if the sequence is Arithmetic or Geometric :}$

$\textsf{If the difference between successive terms is }$ $\textsf{equal then, the terms are in AP}$

• $\sf{ \dfrac{14}{3}- \dfrac{13}{3} = \dfrac{1}{3}}$

• $\sf{ {5}{}- \dfrac{14}{3} = \dfrac{15-14}{3} =\dfrac{1}{3}}$

$\textsf{Since the common difference is same, }$ $\textsf{we can infer that it's an Arithmetic progression}$ $\textsf{with common difference of } \sf \dfrac{1}{3}$

Answer 2

Answer:

Arithmetic with common difference of $\sf \frac{1}{3}$

Step-by-step explanation:

$\textsf{Given sequence}=4, \dfrac{13}{3}, \dfrac{14}{3}, 5, \dfrac{16}{3},...$

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).

$\sf 4\quad \overset{+\frac{1}{3}}{\longrightarrow}\quad\dfrac{13}{3}\quad \overset{+\frac{1}{3}}{\longrightarrow}\quad \dfrac{14}{3}\quad \overset{+\frac{1}{3}}{\longrightarrow}\quad 5\quad \overset{+\frac{1}{3}}{\longrightarrow}\quad \dfrac{16}{3}$

As the difference between consecutive terms is $\sf \frac{1}{3}$ then the sequence is arithmetic with common difference of $\sf \frac{1}{3}$

General form of an arithmetic sequence: $\sf a_n=a+(n-1)d$

where:

• $\sf a_n$ is the nth term
• a is the first term
• d is the common difference between terms

Given:

• a = 4
• $\sf d=\dfrac{1}{3}$

So the formula for the nth term of this sequence is:

$\implies \sf a_n=4+(n-1)\dfrac{1}{3}$

$\implies \sf a_n=\dfrac{1}{3}n+\dfrac{11}{3}$