Question

Determine whether each sequence below is arithmetic, geometric, or neither. Provide support for your conclusions.

SEQUENCE 1: 1/2 , 7/6 , 11/6 , 5/2 .....
SEQUENCE 2: 1/2 , 1/3 , 2/9 , 4/27 .....​​

Answer 1

Answer:

Sequence 1 is an arithmetic sequence because the differences between consecutive terms are constant.

Sequence 2 is a geometric sequence because the ratios between consecutive terms are constant.

Step-by-step explanation:

In an arithmetic sequence, the difference between any two consecutive terms is constant and is called the common difference. In a geometric sequence, each term is obtained by multiplying the preceding term by a fixed, non-zero number known as the common ratio.

To determine whether a sequence is arithmetic, geometric or neither we can examine the differences and ratios between consecutive terms.

Sequence 1

To examine the differences between consecutive terms, subtract a term from the next term:

$a_4-a_3=\dfrac{5}{2}-\dfrac{11}{6}=\dfrac{15}{6}-\dfrac{11}{6}=\dfrac{4}{6}=\dfrac{2}{3}$

$a_3-a_2=\dfrac{11}{6}-\dfrac{7}{6}=\dfrac{4}{6}=\dfrac{2}{3}$

$a_2-a_1=\dfrac{7}{6}-\dfrac{1}{2}=\dfrac{7}{6}-\dfrac{3}{6}=\dfrac{4}{6}=\dfrac{2}{3}$

As the differences between consecutive terms are constant, this sequence is arithmetic.

Sequence 2

To examine the differences between consecutive terms, subtract a term from the next term:

$a_4-a_3=\dfrac{4}{27}-\dfrac{2}{9}=\dfrac{4}{27}-\dfrac{6}{27}=-\dfrac{2}{27}$

$a_3-a_2=\dfrac{2}{9}-\dfrac{1}{3}=\dfrac{2}{9}-\dfrac{3}{9}=-\dfrac{1}{9}$

$a_2-a_1=\dfrac{1}{3}-\dfrac{1}{2}=\dfrac{2}{6}-\dfrac{3}{6}=-\dfrac{1}{6}$

As the differences between consecutive terms are not constant, this sequence is not arithmetic.

Let's check the ratios between consecutive terms by dividing each term by the preceding term:

$\dfrac{a_4}{a_3}=\dfrac{\frac{4}{27}}{\frac{2}{9}}=\dfrac{4}{27}\times \dfrac{9}{2}=\dfrac{36}{54}=\dfrac{2}{3}$

$\dfrac{a_3}{a_2}=\dfrac{\frac{2}{9}}{\frac{1}{3}}=\dfrac{2}{9}\times \dfrac{3}{1}=\dfrac{6}{9}=\dfrac{2}{3}$

$\dfrac{a_2}{a_1}=\dfrac{\frac{1}{3}}{\frac{1}{2}}=\dfrac{1}{3}\times \dfrac{2}{1}=\dfrac{2}{3}$

As the ratios between consecutive terms are constant, the sequence is geometric.