The second term in a geometric sequence is 20. The fourth term in the same sequence is 45/4, or 11.25. What is the common ratio in the sequence?
Thanks for reading :)
Forgot to mention that I'd really appreciate an explanation not just a bare answer.
The second term in a geometric sequence is 20 ... ar = 20
The fourth term 45/4 ... ar^3 = 45/4
Now divide:
ar^3 / ar = (45/4) / 20
r^2 = 45/80 = 9/16
r = ±3/4
To find the common ratio in a geometric sequence, you can use the formula:
\[ \text{common ratio} = \left(\frac{\text{any term}}{\text{previous term}}\right) \]
Given that the second term is 20 and the fourth term is 11.25, we can use these values to find the common ratio:
\[
\text{common ratio} = \left(\frac{\text{fourth term}}{\text{second term}}\right)
\]
Plugging in the values:
\[
\text{common ratio} = \left(\frac{11.25}{20}\right)
\]
Dividing the values:
\[
\text{common ratio} = 0.5625
\]
Therefore, the common ratio in the sequence is 0.5625.
To find the common ratio in a geometric sequence, you need to use the formula for the $n$th term of a geometric sequence:
\[a_n = a_1 \cdot r^{(n-1)},\]
where $a_n$ is the $n$th term, $a_1$ is the first term, and $r$ is the common ratio.
Given that the second term is 20, we can substitute this into the formula:
\[20 = a_1 \cdot r^{(2-1)}.\]
Simplifying, we have:
\[20 = a_1 \cdot r.\]
And given that the fourth term is 11.25, we can substitute this into the formula as well:
\[11.25 = a_1 \cdot r^{(4-1)}.\]
Simplifying, we have:
\[11.25 = a_1 \cdot r^3.\]
Now we have a system of two equations with two unknowns. We can divide the second equation by the first equation to eliminate $a_1$:
\[\frac{11.25}{20} = \frac{a_1 \cdot r^3}{a_1 \cdot r}.\]
Simplifying, we have:
\[\frac{11.25}{20} = r^2.\]
Now we can take the square root of both sides to solve for $r$:
\[r = \sqrt{\frac{11.25}{20}}.\]
Evaluating this expression, we find that the common ratio is:
\[r \approx 0.75.\]
So, the common ratio in the geometric sequence is approximately 0.75.