The first and last term of a geometrical Pregression (G.P.) are 3 and 96 respectively. If the common ratio is 2, find :
(i) 'n' the number of terms of the G.P.
(ii) Sum of the n terms.

Answer

To solve the problem step by step, we will find the number of terms (n) in the geometric progression (G.P.) and then calculate the sum of those n terms.

### Step 1: Identify the given values
- First term (A) = 3
- Last term (An) = 96
- Common ratio (R) = 2

### Step 2: Use the formula for the nth term of a G.P.
The formula for the nth term of a G.P. is given by:
\[ An = A \cdot R^{n-1} \]
Substituting the known values:
\[ 96 = 3 \cdot 2^{n-1} \]

### Step 3: Solve for \( n \)
First, divide both sides by 3:
\[ \frac{96}{3} = 2^{n-1} \]
\[ 32 = 2^{n-1} \]

Next, express 32 as a power of 2:
\[ 32 = 2^5 \]
Now we have:
\[ 2^{n-1} = 2^5 \]

Since the bases are the same, we can equate the exponents:
\[ n - 1 = 5 \]
Adding 1 to both sides gives:
\[ n = 6 \]

### Step 4: Calculate the sum of the n terms
The formula for the sum of the first n terms of a G.P. is:
\[ S_n = \frac{A(R^n - 1)}{R - 1} \]
Substituting the known values:
\[ S_6 = \frac{3(2^6 - 1)}{2 - 1} \]

### Step 5: Calculate \( 2^6 \)
Calculating \( 2^6 \):
\[ 2^6 = 64 \]

### Step 6: Substitute back into the sum formula
Now substitute back into the sum formula:
\[ S_6 = \frac{3(64 - 1)}{1} \]
\[ S_6 = 3 \cdot 63 \]
\[ S_6 = 189 \]

### Final Answers
(i) The number of terms \( n \) is **6**.
(ii) The sum of the n terms is **189**.