The 5th term in a geometric sequence is 120. The 7th term is 30. What are the possible values of the 6th term of the sequence?
(1 point)
60
±60
75
±75
The geometric mean is ±√(30*120) = 60
or, since r^2 = 120/30 = 4,
r = ±2
so the middle term is ±60
Yes, you are correct. Thank you for pointing that out. The correct possible values for the 6th term are ±60. Thank you for your attention to detail.
The general formula for a geometric sequence is given by \(a_n = a_1 \cdot r^{(n-1)}\), where \(a_n\) represents the nth term in the sequence, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.
Given that the 5th term is 120, we can write:
\(a_5 = a_1 \cdot r^{(5-1)} = 120\)
Given that the 7th term is 30, we can write:
\(a_7 = a_1 \cdot r^{(7-1)} = 30\)
Now, we can set up a system of two equations:
1. \(a_1 \cdot r^4 = 120\)
2. \(a_1 \cdot r^6 = 30\)
Dividing the two equations, we get:
\(\frac{a_1 \cdot r^6}{a_1 \cdot r^4} = \frac{30}{120}\)
\(r^2 = \frac{1}{4}\)
\(r = \pm\frac{1}{2}\)
Now, we can find the possible values of the 6th term by substituting the value of \(r\) back into the equation for the 6th term:
\(a_6 = a_1 \cdot r^{(6-1)}\)
If \(r = -\frac{1}{2}\):
\(a_6 = a_1 \cdot \left(-\frac{1}{2}\right)^5 = a_1 \cdot \frac{1}{32} = \frac{a_1}{32}\)
If \(r = \frac{1}{2}\):
\(a_6 = a_1 \cdot \left(\frac{1}{2}\right)^5 = a_1 \cdot \frac{1}{32} = \frac{a_1}{32}\)
Therefore, the possible values of the 6th term in the sequence are \(75\) and \(-75\). So, the correct answer is:
±75