What is the common ratio in the following geometric sequence?

2, \(\frac{1}{2}\), \(\frac{1}{8}\), \(\frac{1}{32}\), . . .
(1 point)
\(\frac{1}{2}\)
4
\(-1.5\)
\(\frac{1}{4}\)

1 answer

To find the common ratio in the geometric sequence, we can divide any term by its preceding term. Let's calculate it using the first two terms:

  1. The first term is \( a_1 = 2 \).
  2. The second term is \( a_2 = \frac{1}{2} \).

Now, we calculate the common ratio \( r \):

\[ r = \frac{a_2}{a_1} = \frac{\frac{1}{2}}{2} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

We can verify by calculating the ratio for the next pair of terms as well:

For \( a_2 = \frac{1}{2} \) and \( a_3 = \frac{1}{8} \):

\[ r = \frac{a_3}{a_2} = \frac{\frac{1}{8}}{\frac{1}{2}} = \frac{1}{8} \times \frac{2}{1} = \frac{2}{8} = \frac{1}{4} \]

And for \( a_3 = \frac{1}{8} \) and \( a_4 = \frac{1}{32} \):

\[ r = \frac{a_4}{a_3} = \frac{\frac{1}{32}}{\frac{1}{8}} = \frac{1}{32} \times \frac{8}{1} = \frac{8}{32} = \frac{1}{4} \]

The common ratio \( r \) is consistent throughout, and its value is \( \frac{1}{4} \).

Thus, the common ratio is:

\(\frac{1}{4}\)