To determine whether the sequence [tex]\(\frac{3}{16}, \frac{1}{4}, \frac{1}{3}, \ldots\)[/tex] is geometric, we must check if there is a common ratio [tex]\( r \)[/tex] between consecutive terms of the sequence.
Let's start by calculating the ratio of the second term to the first term:
[tex]\[
\text{Ratio } r_1 = \frac{a_2}{a_1} = \frac{\frac{1}{4}}{\frac{3}{16}} = \frac{1}{4} \times \frac{16}{3} = \frac{16}{12} = \frac{4}{3} = 1.3333333333333333
\][/tex]
Next, we calculate the ratio of the third term to the second term:
[tex]\[
\text{Ratio } r_2 = \frac{a_3}{a_2} = \frac{\frac{1}{3}}{\frac{1}{4}} = \frac{1}{3} \times 4 = \frac{4}{3} = 1.3333333333333333
\][/tex]
We observe that the ratio of the second term to the first term [tex]\( r_1 \)[/tex] is the same as the ratio of the third term to the second term [tex]\( r_2 \)[/tex]. Both ratios are equal to [tex]\(\frac{4}{3}\)[/tex] or [tex]\( 1.3333333333333333 \)[/tex].
Since the common ratios between consecutive terms are equal, the sequence [tex]\(\frac{3}{16}, \frac{1}{4}, \frac{1}{3}, \ldots\)[/tex] is indeed a geometric sequence, and the common ratio is:
[tex]\[
r = 1.3333333333333333
\][/tex]
