Question

find the 6th term of this geometric sequence.

(3, -12, 48, \dots)

(a_6 = ?)

hint: (a_n = a_1 \cdot r^{(n-1)})

Explanation:

Identify the first term and target term

We are given the geometric sequence:

$$3, -12, 48, \dots$$

The first term of the sequence is:

$$a_1 = 3$$

We want to find the 6th term, which means:

$$n = 6$$

Calculate the common ratio

To find the common ratio \(r\), we divide any term by its preceding term:

$$r = \frac{a_2}{a_1} = \frac{-12}{3} = -4$$

We can verify this with the next term:

$$\frac{a_3}{a_2} = \frac{48}{-12} = -4$$

Thus, the common ratio is \(r = -4\).

Apply the geometric sequence formula

The formula for the \(n\)-th term of a geometric sequence is:

$$a_n = a_1 \cdot r^{(n-1)}$$

Substituting our known values \(a_1 = 3\), \(r = -4\), and \(n = 6\):

$$a_6 = 3 \cdot (-4)^{(6-1)}$$

$$a_6 = 3 \cdot (-4)^5$$

Compute the final value

First, calculate the exponent:

$$(-4)^5 = -1024$$

Now, multiply by the first term:

$$a_6 = 3 \cdot (-1024) = -3072$$

Answer:

Find the 6th term of this geometric sequence.

\(3, -12, 48, \dots\)

\(a_6 =\) <blank>-3072</blank>