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

Options

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

Answer

To find the nth term of the sequence \( 25, -125, 625, -3125, \ldots \), we can follow these steps:

### Step 1: Identify the first term and the common ratio
The first term \( a \) of the sequence is \( 25 \). To find the common ratio \( r \), we can divide the second term by the first term:
\[ r = \frac{-125}{25} = -5 \]

### Step 2: Confirm the common ratio
We can check the common ratio for the subsequent terms:
- From the second term to the third term:
\[ \frac{625}{-125} = -5 \]
- From the third term to the fourth term:
\[ \frac{-3125}{625} = -5 \]
Thus, the common ratio \( r \) is confirmed to be \( -5 \).

### Step 3: Write the formula for the nth term of a geometric progression
The nth term \( T_n \) of a geometric progression can be calculated using the formula:
\[ T_n = a \cdot r^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio.

### Step 4: Substitute the values into the formula
Substituting \( a = 25 \) and \( r = -5 \) into the formula gives:
\[ T_n = 25 \cdot (-5)^{n-1} \]

### Step 5: Simplify the expression
We can express \( (-5)^{n-1} \) as:
\[ T_n = 25 \cdot (-1)^{n-1} \cdot 5^{n-1} \]
This can be rewritten as:
\[ T_n = (-1)^{n-1} \cdot 25 \cdot 5^{n-1} \]
Since \( 25 = 5^2 \), we can further simplify:
\[ T_n = (-1)^{n-1} \cdot 5^{n-1 + 2} = (-1)^{n-1} \cdot 5^{n+1} \]

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