If `3^(2sin2alpha-1),14` and `3^(4-2sin2alpha)` be first three terms of an A.P for some `alpha`, then find `6^(th)` term of this A.P.

Answer

To solve the problem, we need to determine the sixth term of an arithmetic progression (A.P.) whose first three terms are given as \(3^{(2\sin 2\alpha - 1)}, 14,\) and \(3^{(4 - 2\sin 2\alpha)}\).

### Step 1: Identify the terms of the A.P.
Let:
- \(a_1 = 3^{(2\sin 2\alpha - 1)}\)
- \(a_2 = 14\)
- \(a_3 = 3^{(4 - 2\sin 2\alpha)}\)

### Step 2: Use the property of A.P.
In an A.P., the middle term is the average of the other two terms. Therefore, we can set up the equation:
\[ a_2 = \frac{a_1 + a_3}{2} \]
Substituting the values we have:
\[ 14 = \frac{3^{(2\sin 2\alpha - 1)} + 3^{(4 - 2\sin 2\alpha)}}{2} \]

### Step 3: Multiply both sides by 2 to eliminate the fraction.
\[ 28 = 3^{(2\sin 2\alpha - 1)} + 3^{(4 - 2\sin 2\alpha)} \]

### Step 4: Simplify the equation.
Let \(x = 3^{(2\sin 2\alpha)}\). Then, we can rewrite the terms:
\[ 3^{(2\sin 2\alpha - 1)} = \frac{x}{3} \quad \text{and} \quad 3^{(4 - 2\sin 2\alpha)} = \frac{81}{x} \]
Substituting these into the equation gives:
\[ 28 = \frac{x}{3} + \frac{81}{x} \]

### Step 5: Multiply through by \(3x\) to eliminate the denominators.
\[ 28 \cdot 3x = x^2 + 243 \]
This simplifies to:
\[ 84x = x^2 + 243 \]

### Step 6: Rearrange into standard quadratic form.
\[ x^2 - 84x + 243 = 0 \]

### Step 7: Solve the quadratic equation using the quadratic formula.
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -84\), and \(c = 243\):
\[ x = \frac{84 \pm \sqrt{(-84)^2 - 4 \cdot 1 \cdot 243}}{2 \cdot 1} \]
Calculating the discriminant:
\[ (-84)^2 = 7056 \quad \text{and} \quad 4 \cdot 1 \cdot 243 = 972 \]
Thus:
\[ \sqrt{7056 - 972} = \sqrt{6084} = 78 \]
Now substituting back:
\[ x = \frac{84 \pm 78}{2} \]
Calculating the two possible values:
1. \(x = \frac{162}{2} = 81\)
2. \(x = \frac{6}{2} = 3\)

### Step 8: Find the corresponding values of \(\sin 2\alpha\).
Since \(x = 3^{(2\sin 2\alpha)}\), we have:
1. If \(x = 81\), then \(2\sin 2\alpha = 4 \Rightarrow \sin 2\alpha = 2\) (not possible).
2. If \(x = 3\), then \(2\sin 2\alpha = 1 \Rightarrow \sin 2\alpha = \frac{1}{2} \Rightarrow 2\alpha = \frac{\pi}{6} \Rightarrow \alpha = \frac{\pi}{12}\).

### Step 9: Calculate the terms of the A.P.
Substituting \(\alpha = \frac{\pi}{12}\):
- \(a_1 = 3^{(2\sin 2\alpha - 1)} = 3^{(0 - 1)} = \frac{1}{3}\)
- \(a_2 = 14\)
- \(a_3 = 3^{(4 - 2\sin 2\alpha)} = 3^{(4 - 1)} = 27\)

### Step 10: Find the common difference \(d\).
The common difference \(d\) is:
\[ d = a_2 - a_1 = 14 - \frac{1}{3} = \frac{42 - 1}{3} = \frac{41}{3} \]

### Step 11: Calculate the sixth term of the A.P.
Using the formula for the \(n\)-th term of an A.P.:
\[ a_n = a_1 + (n-1)d \]
For \(n = 6\):
\[ a_6 = \frac{1}{3} + 5 \cdot \frac{41}{3} = \frac{1 + 205}{3} = \frac{206}{3} \]

### Final Answer:
Thus, the sixth term of the A.P. is \(\frac{206}{3}\).

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