In a geometric series, suppose g2 = 2 and g6 = 162. What is gngn–1?

Sorry last part should be

What is the ratio of the geometric sequence?

To determine the value of gngn–1, we need to identify the common ratio (r) of the geometric series first.

The geometric series is typically expressed as: gn = g1 * r^(n-1), where g1 is the first term of the series, r is the common ratio, and n represents the term number.

Given g2 = 2 and g6 = 162, we can write the equations as:

g2 = g1 * r^(2-1)
2 = g1 * r

g6 = g1 * r^(6-1)
162 = g1 * r^5

Now, let's solve these two equations simultaneously to find the values of g1 and r. We'll divide the second equation by the first equation to eliminate g1:

(162 / 2) = (g1 * r^5) / (g1 * r)
81 = r^4

Taking the fourth root of both sides, we get:

r = ∛(81)
r = 3

Now that we know the common ratio (r = 3), we can substitute it back into either of the original equations to find g1. Let's use the first equation:

2 = g1 * 3

Dividing both sides by 3:

2/3 = g1

Therefore, g1 = 2/3.

Lastly, to find gngn–1, we substitute n into the formula gn = g1 * r^(n-1):

gngn–1 = (2/3) * (3)^(n-1)

Since the value of n is not provided, we cannot determine the exact value of gngn–1 without knowing that information.