To solve this problem, we use the formula for the general term $a_n$ of a geometric sequence:

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

Given:

• $a_1 = 5$ (the first term)
• $r = -3$ (the common ratio)

Step 1: Write the formula for $a_n$:

Substitute $a_1 = 5$ and $r = -3$ into the formula:

$a_n = 5 \cdot (-3)^{n-1}$

Step 2: Find $a_5$ (the 5th term):

To find $a_5$, set $n = 5$:

$a_5 = 5 \cdot (-3)^{5-1}$ Simplify the exponent:

$a_5 = 5 \cdot (-3)^4$

Now calculate $(-3)^4$: $(-3)^4 = 81$

Thus: $a_5 = 5 \cdot 81 = 405$

Final Answers:

1. General formula: $a_n = 5 \cdot (-3)^{n-1}$
2. $a_5 = 405$

Let me know if you would like more explanation or further clarifications!

Related Questions:

1. What is the 6th term ($a_6$) of this sequence?
2. How does the sign of the common ratio $r$ affect the sequence?
3. What happens if $r$ is positive instead of negative?
4. Derive the sum of the first 5 terms of this geometric sequence.
5. What is the $n$-th term if $r = -2$ and $a_1 = 4$?

Tip:

In geometric sequences, alternating signs occur if the common ratio $r$ is negative.