The 5th and 12th terms of a G.P. are 32 and 4096 respectively . Find the nth term of the G.P. :

Options

A. `2^n`
B. `n^2`
C. `2n^2`
D. none of these

Answer

To solve the problem step by step, we need to find the nth term of a geometric progression (G.P.) given the 5th and 12th terms.

### Step 1: Understand the terms of a G.P.
The nth term of a G.P. can be expressed as:
\[ A_n = A \cdot R^{n-1} \]
where \( A \) is the first term and \( R \) is the common ratio.

### Step 2: Write the equations for the given terms
We know:
- The 5th term \( A_5 = A \cdot R^{4} = 32 \) (Equation 1)
- The 12th term \( A_{12} = A \cdot R^{11} = 4096 \) (Equation 2)

### Step 3: Divide Equation 2 by Equation 1
To eliminate \( A \), we divide Equation 2 by Equation 1:
\[ \frac{A \cdot R^{11}}{A \cdot R^{4}} = \frac{4096}{32} \]
This simplifies to:
\[ R^{7} = \frac{4096}{32} \]

### Step 4: Simplify the right side
Calculating the right side:
\[ \frac{4096}{32} = 128 \]
Thus, we have:
\[ R^{7} = 128 \]

### Step 5: Express 128 as a power of 2
We can express 128 as:
\[ 128 = 2^{7} \]
So, we have:
\[ R^{7} = 2^{7} \]

### Step 6: Solve for \( R \)
Since the bases are the same, we can equate the exponents:
\[ R = 2 \]

### Step 7: Substitute \( R \) back to find \( A \)
Now that we have \( R \), we can substitute it back into Equation 1 to find \( A \):
\[ A \cdot (2^{4}) = 32 \]
This simplifies to:
\[ A \cdot 16 = 32 \]
Thus:
\[ A = \frac{32}{16} = 2 \]

### Step 8: Write the nth term of the G.P.
Now that we have both \( A \) and \( R \), we can write the nth term:
\[ A_n = A \cdot R^{n-1} = 2 \cdot 2^{n-1} \]
This simplifies to:
\[ A_n = 2^{1 + (n-1)} = 2^{n} \]

### Final Answer
The nth term of the G.P. is:
\[ A_n = 2^{n} \]

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