Question

Use the recursively defined geometric sequence a1=5/6, an=4an−1 and find the common ratio.(1 point)
Responses

Answer 1

In a geometric sequence, each term is obtained by multiplying the previous term by a constant called the common ratio.

In this case, we have:

- \( a_1 = \frac{5}{6} \)
- \( a_n = 4a_{n-1} \)

To find the common ratio, we can express \( a_2 \) in terms of \( a_1 \):

\[
a_2 = 4a_1 = 4 \left(\frac{5}{6}\right) = \frac{20}{6} = \frac{10}{3}
\]

Next, we can find \( a_3 \):

\[
a_3 = 4a_2 = 4 \left(\frac{10}{3}\right) = \frac{40}{3}
\]

Now we can calculate the ratio between consecutive terms:

The ratio \( r \) between \( a_2 \) and \( a_1 \) is:

\[
r = \frac{a_2}{a_1} = \frac{\frac{10}{3}}{\frac{5}{6}} = \frac{10}{3} \cdot \frac{6}{5} = \frac{60}{15} = 4
\]

The ratio \( r \) between \( a_3 \) and \( a_2 \) is:

\[
r = \frac{a_3}{a_2} = \frac{\frac{40}{3}}{\frac{10}{3}} = \frac{40}{10} = 4
\]

Since this ratio is consistent between all consecutive terms, the common ratio of the geometric sequence is:

**Common Ratio \( r = 4 \)**