Find the sum of G.P. : 3,6,12, . . . . . . . . ., 1536.

Answer

To find the sum of the geometric progression (G.P.) given by the terms 3, 6, 12, ..., up to 1536, we can follow these steps:

### Step 1: Identify the first term and common ratio
The first term \( a \) of the G.P. is:
\[ a = 3 \]
The common ratio \( r \) can be calculated as:
\[ r = \frac{6}{3} = 2 \quad \text{(also, } \frac{12}{6} = 2\text{)} \]

### Step 2: Write the formula for the nth term of a G.P.
The nth term \( T_n \) of a G.P. is given by:
\[ T_n = a \cdot r^{n-1} \]
We know that the last term \( T_n \) is 1536, so we set up the equation:
\[ T_n = 3 \cdot 2^{n-1} = 1536 \]

### Step 3: Solve for \( n \)
To find \( n \), we rearrange the equation:
\[ 2^{n-1} = \frac{1536}{3} = 512 \]
Next, we express 512 as a power of 2:
\[ 512 = 2^9 \]
Thus, we have:
\[ n - 1 = 9 \implies n = 10 \]

### Step 4: Calculate the sum of the first \( n \) terms of the G.P.
The sum \( S_n \) of the first \( n \) terms of a G.P. is given by:
\[ S_n = a \cdot \frac{r^n - 1}{r - 1} \]
Substituting the known values:
\[ S_{10} = 3 \cdot \frac{2^{10} - 1}{2 - 1} \]
Calculating \( 2^{10} \):
\[ 2^{10} = 1024 \]
Now substituting back:
\[ S_{10} = 3 \cdot \frac{1024 - 1}{1} = 3 \cdot 1023 = 3069 \]

### Final Answer
Thus, the sum of the G.P. is:
\[ \boxed{3069} \]