paynelove19
Answered

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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?

Sagot :

DᴀʀᴋPᴀʀᴀᴅᴏx

[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]

[tex] \textbf{Let's see if the sequence is Arithmetic or Geometric :} [/tex]

[tex] \textsf{If the difference between successive terms is } [/tex] [tex] \textsf{equal then, the terms are in AP} [/tex]

  • [tex] \sf{ \dfrac{14}{3}- \dfrac{13}{3} = \dfrac{1}{3}} [/tex]

  • [tex] \sf{ {5}{}- \dfrac{14}{3} = \dfrac{15-14}{3} =\dfrac{1}{3}} [/tex]

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

semsee45

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]

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