The nth term of a G.P. is `3cdot(-2)^(n).` Find the sum of its 7 terms.

Explanation

To find the sum of the first 7 terms of the given geometric progression (G.P.) where the nth term is given by \( a_n = 3 \cdot (-2)^n \), we will follow these steps:

### Step 1: Identify the first term and the common ratio
The first term \( a_1 \) can be found by substituting \( n = 1 \) into the formula for \( a_n \):
\[ a_1 = 3 \cdot (-2)^1 = 3 \cdot (-2) = -6 \]

The second term \( a_2 \) can be found by substituting \( n = 2 \):
\[ a_2 = 3 \cdot (-2)^2 = 3 \cdot 4 = 12 \]

Now, we can find the common ratio \( r \):
\[ r = \frac{a_2}{a_1} = \frac{12}{-6} = -2 \]

### Step 2: Use the formula for the sum of the first n terms of a G.P.
The formula for the sum of the first \( n \) terms of a G.P. is given by:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

For our case:
- \( a = -6 \)
- \( r = -2 \)
- \( n = 7 \)

### Step 3: Substitute the values into the formula
Now, substituting these values into the formula for \( S_7 \):
\[ S_7 = -6 \cdot \frac{1 - (-2)^7}{1 - (-2)} \]

### Step 4: Calculate \( (-2)^7 \)
Calculating \( (-2)^7 \):
\[ (-2)^7 = -128 \]

### Step 5: Substitute back into the sum formula
Now substituting back:
\[ S_7 = -6 \cdot \frac{1 - (-128)}{1 + 2} \]
\[ S_7 = -6 \cdot \frac{1 + 128}{3} \]
\[ S_7 = -6 \cdot \frac{129}{3} \]

### Step 6: Simplify the expression
Calculating \( \frac{129}{3} \):
\[ \frac{129}{3} = 43 \]
Thus,
\[ S_7 = -6 \cdot 43 = -258 \]

### Final Answer
The sum of the first 7 terms of the G.P. is:
\[ \boxed{-258} \]