The common ratio, last term and sum of n terms of a G.P. are 2, 128 and 255 respectively. Find the value of n.

Explanation

To solve the problem step by step, we will use the properties of a geometric progression (G.P.) and the formulas for the nth term and the sum of n terms.

### Given:
- Common ratio \( r = 2 \)
- Last term \( a_n = 128 \)
- Sum of n terms \( S_n = 255 \)

### Step 1: Use the formula for the nth term of a G.P.
The nth term of a G.P. is given by the formula:
\[ a_n = a \cdot r^{n-1} \]
Substituting the known values:
\[ 128 = a \cdot 2^{n-1} \quad \text{(Equation 1)} \]

### Step 2: Use the formula for the sum of n terms of a G.P.
The sum of the first n terms of a G.P. is given by the formula:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
Substituting the known values:
\[ 255 = \frac{a(2^n - 1)}{2 - 1} \]
This simplifies to:
\[ 255 = a(2^n - 1) \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
From Equation 1, we can express \( a \) in terms of \( n \):
\[ a = \frac{128}{2^{n-1}} \]

### Step 4: Substitute \( a \) into Equation 2
Substituting \( a \) into Equation 2:
\[ 255 = \left(\frac{128}{2^{n-1}}\right)(2^n - 1) \]
This simplifies to:
\[ 255 = 128 \cdot \frac{(2^n - 1)}{2^{n-1}} \]
Multiplying both sides by \( 2^{n-1} \):
\[ 255 \cdot 2^{n-1} = 128(2^n - 1) \]
Expanding the right side:
\[ 255 \cdot 2^{n-1} = 128 \cdot 2^n - 128 \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 128 \cdot 2^n - 255 \cdot 2^{n-1} - 128 = 0 \]
Factoring out \( 2^{n-1} \):
\[ 128 \cdot 2^{n-1} - 255 \cdot 2^{n-1} - 128 = 0 \]
This simplifies to:
\[ (128 - 255) \cdot 2^{n-1} = 128 \]
\[ -127 \cdot 2^{n-1} = 128 \]
Dividing both sides by -127:
\[ 2^{n-1} = -\frac{128}{127} \]
This is incorrect since \( 2^{n-1} \) cannot be negative.

### Step 6: Correct the equation
Revisiting the equation:
\[ 255 \cdot 2^{n-1} = 128 \cdot 2^n - 128 \]
Rearranging gives:
\[ 128 \cdot 2^n - 255 \cdot 2^{n-1} = 128 \]
Factoring out \( 2^{n-1} \):
\[ 2^{n-1}(128 \cdot 2 - 255) = 128 \]
\[ 2^{n-1}(256 - 255) = 128 \]
\[ 2^{n-1} = 128 \]
Since \( 128 = 2^7 \):
\[ 2^{n-1} = 2^7 \]
Thus, \( n - 1 = 7 \) leading to:
\[ n = 8 \]

### Final Answer:
The value of \( n \) is \( 8 \).