If `3^(2sin2x -1), 14, 3^(4 -2sin2x)` form first three terms of an A.P then find the value of `1+ sin2x+ sin^2(2x)+........`

Answer

To solve the problem, we need to determine the value of \(1 + \sin 2x + \sin^2(2x) + \ldots\) given that the terms \(3^{2\sin 2x - 1}\), \(14\), and \(3^{4 - 2\sin 2x}\) form an arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Understanding the A.P. Condition**:
Since the three terms form an A.P., the middle term is the average of the other two terms. Therefore, we can write:
\[ 14 = \frac{3^{2\sin 2x - 1} + 3^{4 - 2\sin 2x}}{2} \]
Multiplying both sides by 2 gives:
\[ 28 = 3^{2\sin 2x - 1} + 3^{4 - 2\sin 2x} \]

2. **Rewriting the Exponential Terms**:
We can express the terms in a more manageable form:
Let \(y = 2\sin 2x\). Then, we can rewrite the equation as:
\[ 28 = 3^{y - 1} + 3^{4 - y} \]
This simplifies to:
\[ 28 = \frac{3^y}{3} + \frac{81}{3^y} \]
Multiplying through by 3 gives:
\[ 84 = 3^y + \frac{81}{3^y} \]

3. **Multiplying by \(3^y\)**:
To eliminate the fraction, we multiply through by \(3^y\):
\[ 84 \cdot 3^y = 3^{2y} + 81 \]
Rearranging gives us a quadratic equation:
\[ 3^{2y} - 84 \cdot 3^y + 81 = 0 \]

4. **Letting \(z = 3^y\)**:
Substituting \(z = 3^y\), we have:
\[ z^2 - 84z + 81 = 0 \]

5. **Solving the Quadratic Equation**:
Using the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ z = \frac{84 \pm \sqrt{84^2 - 4 \cdot 1 \cdot 81}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 84^2 = 7056, \quad 4 \cdot 81 = 324 \quad \Rightarrow \quad 7056 - 324 = 6732 \]
Thus:
\[ z = \frac{84 \pm \sqrt{6732}}{2} \]

6. **Finding Roots**:
We can approximate \(\sqrt{6732} \approx 82\) (since \(82^2 = 6724\)), so:
\[ z \approx \frac{84 \pm 82}{2} \]
This gives us two possible values:
\[ z_1 \approx 83, \quad z_2 \approx 1 \]

7. **Finding \( \sin 2x \)**:
Since \(z = 3^{2\sin 2x}\), we can find \(2\sin 2x\):
- If \(z = 83\), then \(2\sin 2x = \log_3(83)\) (not useful as \(\sin\) cannot exceed 1).
- If \(z = 1\), then \(2\sin 2x = 0\) which gives \(\sin 2x = 0\).

8. **Calculating the Infinite Series**:
Now we need to evaluate:
\[ 1 + \sin 2x + \sin^2(2x) + \ldots \]
This is a geometric series with first term \(a = 1\) and common ratio \(r = \sin 2x\). The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Here, \(a = 1\) and \(r = \sin 2x = 0\):
\[ S = \frac{1}{1 - 0} = 1 \]

### Final Answer:
Thus, the value of \(1 + \sin 2x + \sin^2(2x) + \ldots\) is:
\[ \boxed{2} \]