The numbers `3^(2sin2alpha-1),14and3^(4-2sin2alpha)` form first three terms of A.P., its fifth term is

Options

A. `-25`
B. `-12`
C. 40
D. 53

Explanation

To solve the problem, we need to find the fifth term of the arithmetic progression (A.P.) formed by the numbers \(3^{(2\sin 2\alpha - 1)}, 14,\) and \(3^{(4 - 2\sin 2\alpha)}\).

### Step-by-Step Solution:

1. **Identify the Terms in A.P.**
The first three terms of the A.P. are:
- \(a_1 = 3^{(2\sin 2\alpha - 1)}\)
- \(a_2 = 14\)
- \(a_3 = 3^{(4 - 2\sin 2\alpha)}\)

2. **Use the Property of A.P.**
For three numbers to be in A.P., the middle term must be the average of the other two terms:
\[ 2a_2 = a_1 + a_3 \]
Substituting the values, we get:
\[ 2 \cdot 14 = 3^{(2\sin 2\alpha - 1)} + 3^{(4 - 2\sin 2\alpha)} \]
Simplifying this gives:
\[ 28 = 3^{(2\sin 2\alpha - 1)} + 3^{(4 - 2\sin 2\alpha)} \]

3. **Let \(A = 3^{(2\sin 2\alpha)}\)**
We can rewrite the equation using \(A\):
\[ 28 = \frac{A}{3} + 3^4 \cdot \frac{1}{A} \]
This simplifies to:
\[ 28 = \frac{A}{3} + \frac{81}{A} \]

4. **Multiply through by \(3A\)**
To eliminate the fractions:
\[ 28 \cdot 3A = A^2 + 243 \]
This simplifies to:
\[ 84A = A^2 + 243 \]
Rearranging gives:
\[ A^2 - 84A + 243 = 0 \]

5. **Solve the Quadratic Equation**
We can use the quadratic formula \(A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ A = \frac{84 \pm \sqrt{(-84)^2 - 4 \cdot 1 \cdot 243}}{2 \cdot 1} \]
\[ A = \frac{84 \pm \sqrt{7056 - 972}}{2} \]
\[ A = \frac{84 \pm \sqrt{6084}}{2} \]
\[ A = \frac{84 \pm 78}{2} \]
This gives us two possible values for \(A\):
\[ A = \frac{162}{2} = 81 \quad \text{or} \quad A = \frac{6}{2} = 3 \]

6. **Find \( \sin 2\alpha \)**
Since \(A = 3^{(2\sin 2\alpha)}\):
- If \(A = 81\), then \(2\sin 2\alpha = 4\) (not possible since \(\sin\) cannot exceed 1).
- If \(A = 3\), then \(2\sin 2\alpha = 2\) which gives \(\sin 2\alpha = 1\).

7. **Determine \(\alpha\)**
From \(\sin 2\alpha = 1\), we have:
\[ 2\alpha = \frac{\pi}{2} \implies \alpha = \frac{\pi}{4} \]

8. **Calculate the Terms**
Now substituting \(\alpha = \frac{\pi}{4}\):
- \(a_1 = 3^{(2\sin 2\cdot \frac{\pi}{4} - 1)} = 3^{(2 - 1)} = 3^1 = 3\)
- \(a_3 = 3^{(4 - 2\sin 2\cdot \frac{\pi}{4})} = 3^{(4 - 2)} = 3^2 = 9\)

Therefore, the terms of the A.P. are \(3, 14, 9\).

9. **Find the Common Difference \(d\)**
The common difference \(d\) is:
\[ d = 14 - 3 = 11 \]

10. **Calculate the Fifth Term**
The fifth term \(a_5\) of the A.P. is given by:
\[ a_5 = a_1 + 4d = 3 + 4 \cdot 11 = 3 + 44 = 47 \]

### Final Answer:
The fifth term of the A.P. is \(47\).