What is the `n^(th)` term of the sequence `25,-125,625,-3125,...?`

Options

A. `(-5)^(2n-1)`
B. `(-1)^(2n)5^(n+1)`
C. `(-1)^(2n-1)5^(n+1)`
D. `(-1)^(n-1)5^(n+1)`

Explanation

To find the \( n^{th} \) term of the sequence \( 25, -125, 625, -3125, \ldots \), we can follow these steps:

### Step 1: Identify the first term and common ratio
The first term of the sequence, denoted as \( a \), is:
\[ a = 25 \]

Next, we need to find the common ratio \( r \). We can calculate it by dividing the second term by the first term:
\[ r = \frac{-125}{25} = -5 \]

### Step 2: Confirm the common ratio
To ensure that the sequence is geometric, we can check the ratio of other consecutive terms:
\[ \frac{625}{-125} = -5 \quad \text{and} \quad \frac{-3125}{625} = -5 \]
Since the ratio is consistent, the common ratio \( r \) is confirmed to be \( -5 \).

### Step 3: Write the formula for the \( n^{th} \) term of a geometric progression
The formula for the \( n^{th} \) term of a geometric progression is given by:
\[ T_n = a \cdot r^{n-1} \]
Substituting the values we found:
\[ T_n = 25 \cdot (-5)^{n-1} \]

### Step 4: Simplify the expression
We can express \( 25 \) as \( 5^2 \):
\[ T_n = 5^2 \cdot (-5)^{n-1} \]
This can be rewritten as:
\[ T_n = 5^2 \cdot (-1)^{n-1} \cdot 5^{n-1} \]
Combining the powers of \( 5 \):
\[ T_n = (-1)^{n-1} \cdot 5^{2 + (n-1)} = (-1)^{n-1} \cdot 5^{n+1} \]

### Final Result
Thus, the \( n^{th} \) term of the sequence is:
\[ T_n = (-1)^{n-1} \cdot 5^{n+1} \]