The `n^(th)` term of a GP is 128 and the sum of its n terms is 255. If its common ratio is 2 then its first term is.

Options

A. 1
B. 3
C. 5
D. 7

Answer

To find the first term of the geometric progression (GP), we will follow these steps:

### Step 1: Use the nth term formula of a GP
The nth term of a GP is given by the formula:
\[ a_n = a \cdot r^{n-1} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the term number. We know that the nth term is 128, so we can write:
\[ a \cdot r^{n-1} = 128 \]

### Step 2: Substitute the common ratio
Given that the common ratio \( r = 2 \), we substitute \( r \) into the equation:
\[ a \cdot 2^{n-1} = 128 \]

### Step 3: Solve for \( a \cdot 2^{n-1} \)
To isolate \( a \), we can rearrange the equation:
\[ a \cdot 2^{n-1} = 128 \]
This can be rewritten as:
\[ a = \frac{128}{2^{n-1}} \]

### Step 4: Use the sum of the first n terms formula
The sum of the first \( n \) terms of a GP is given by:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
We know that the sum of the first \( n \) terms is 255, so:
\[ \frac{a(2^n - 1)}{2 - 1} = 255 \]
This simplifies to:
\[ a(2^n - 1) = 255 \]

### Step 5: Substitute \( a \) from Step 3 into the sum equation
Now, we substitute \( a \) from Step 3 into the sum equation:
\[ \frac{128}{2^{n-1}}(2^n - 1) = 255 \]

### Step 6: Simplify the equation
Multiplying both sides by \( 2^{n-1} \) gives:
\[ 128(2^n - 1) = 255 \cdot 2^{n-1} \]
This can be simplified to:
\[ 128 \cdot 2^n - 128 = 255 \cdot 2^{n-1} \]

### Step 7: Factor out \( 2^{n-1} \)
We can express \( 2^n \) as \( 2 \cdot 2^{n-1} \):
\[ 128 \cdot 2 \cdot 2^{n-1} - 128 = 255 \cdot 2^{n-1} \]
This simplifies to:
\[ 256 \cdot 2^{n-1} - 128 = 255 \cdot 2^{n-1} \]

### Step 8: Rearrange the equation
Rearranging gives:
\[ 256 \cdot 2^{n-1} - 255 \cdot 2^{n-1} = 128 \]
This simplifies to:
\[ (256 - 255) \cdot 2^{n-1} = 128 \]
Thus:
\[ 2^{n-1} = 128 \]

### Step 9: Solve for \( n \)
Since \( 128 = 2^7 \), we have:
\[ 2^{n-1} = 2^7 \implies n - 1 = 7 \implies n = 8 \]

### Step 10: Find \( a \)
Now we can substitute \( n \) back into the equation for \( a \):
\[ a = \frac{128}{2^{7}} = \frac{128}{128} = 1 \]

### Final Answer
The first term \( a \) is:
\[ \boxed{1} \]