In a sequence of (4n + 1) terms the first (2n + 1) terms are in AP whose common difference is 2, and the last (2n + 1) terms are in GP whose common ratio is 0.5. If the middle terms of the AP and GP are equal, then the middle term of the sequence is

Options

A. `(n.2^(n+1))/(2^(n)-1)`
B. `(n.2^(n+1))/(2^(2n-1))`
C. `n2^(n)`
D. None of these

Explanation

To solve the problem step by step, we will break it down into manageable parts.

### Step 1: Understand the sequence structure
The sequence consists of \(4n + 1\) terms. The first \(2n + 1\) terms form an Arithmetic Progression (AP) with a common difference of 2, and the last \(2n + 1\) terms form a Geometric Progression (GP) with a common ratio of 0.5.

### Step 2: Define the terms of the AP
Let the first term of the AP be \(A\). The terms of the AP can be expressed as:
- First term: \(A\)
- Second term: \(A + 2\)
- Third term: \(A + 4\)
- ...
- Last term (which is the \((2n + 1)\)-th term): \(A + 4n\)

Thus, the \(k\)-th term of the AP can be given by:
\[ T_k = A + 2(k - 1) \quad \text{for } k = 1, 2, \ldots, 2n + 1 \]

### Step 3: Find the middle term of the AP
The middle term of the AP (which is the \((n + 1)\)-th term) is:
\[ \text{Middle term of AP} = A + 2n \]

### Step 4: Define the terms of the GP
The first term of the GP starts after the AP, which is \(A + 4n\). The terms of the GP can be expressed as:
- First term: \(A + 4n\)
- Second term: \(\frac{A + 4n}{2}\)
- Third term: \(\frac{A + 4n}{4}\)
- ...
- Last term (which is the \((2n + 1)\)-th term): \(\frac{A + 4n}{2^n}\)

Thus, the \(k\)-th term of the GP can be given by:
\[ T_k = (A + 4n) \left(\frac{1}{2}\right)^{k - 1} \quad \text{for } k = 1, 2, \ldots, 2n + 1 \]

### Step 5: Find the middle term of the GP
The middle term of the GP (which is also the \((n + 1)\)-th term) is:
\[ \text{Middle term of GP} = (A + 4n) \left(\frac{1}{2}\right)^{n} \]

### Step 6: Set the middle terms equal
According to the problem, the middle terms of the AP and GP are equal:
\[ A + 2n = (A + 4n) \left(\frac{1}{2}\right)^{n} \]

### Step 7: Solve for \(A\)
Rearranging the equation gives:
\[ A + 2n = \frac{A + 4n}{2^n} \]
Multiplying both sides by \(2^n\):
\[ 2^n(A + 2n) = A + 4n \]
Expanding and rearranging:
\[ 2^n A + 2^{n + 1} n = A + 4n \]
\[ (2^n - 1)A = 4n - 2^{n + 1} n \]
Thus,
\[ A = \frac{4n - 2^{n + 1} n}{2^n - 1} \]

### Step 8: Find the middle term of the sequence
The middle term of the entire sequence (which is the \((2n + 1)\)-th term) is:
\[ \text{Middle term of sequence} = A + 4n \]
Substituting the value of \(A\):
\[ \text{Middle term of sequence} = \frac{4n - 2^{n + 1} n}{2^n - 1} + 4n \]
Combining the terms:
\[ \text{Middle term of sequence} = \frac{4n - 2^{n + 1} n + 4n(2^n - 1)}{2^n - 1} \]
\[ = \frac{4n - 2^{n + 1} n + 4n \cdot 2^n - 4n}{2^n - 1} \]
\[ = \frac{4n \cdot 2^n - 2^{n + 1} n}{2^n - 1} \]

### Final Result
The middle term of the sequence is:
\[ \frac{n(2^{n + 1} - 4)}{2^n - 1} \]