Key Concepts and Formulas

• Arithmetic Progression (A.P.): A sequence where the difference between consecutive terms is constant (common difference, $d$). The property $2a_2 = a_1 + a_3$ holds for the first three terms $a_1, a_2, a_3$.
• N-th Term of an A.P.: The $n$-th term is given by $T_n = a_1 + (n-1)d$.
• Exponential Properties: $a^{m-n} = \frac{a^m}{a^n}$ and $a^m = a^n \implies m=n$ for $a>0, a \neq 1$.
• Trigonometric Range: For any real angle $\theta$, $-1 \le \sin \theta \le 1$.

Step-by-Step Solution

Step 1: Apply the A.P. Property We are given that $3^{2\sin 2\alpha - 1}$, 14, and $3^{4 - 2\sin 2\alpha}$ are the first three terms of an A.P. Let these terms be $a_1, a_2,$ and $a_3$ respectively. The fundamental property of an A.P. states that the middle term is the arithmetic mean of the first and third terms, i.e., $2a_2 = a_1 + a_3$. Substituting the given terms: $2 \times 14 = 3^{2\sin 2\alpha - 1} + 3^{4 - 2\sin 2\alpha}$ $28 = 3^{2\sin 2\alpha - 1} + 3^{4 - 2\sin 2\alpha}$

Step 2: Simplify the Exponential Equation using Substitution To solve this equation, let's simplify the exponents. Let $x = 2\sin 2\alpha$. The equation becomes: $28 = 3^{x - 1} + 3^{4 - x}$ Using the property $a^{m-n} = \frac{a^m}{a^n}$: $3^{x - 1} = \frac{3^x}{3^1} = \frac{3^x}{3}$ $3^{4 - x} = \frac{3^4}{3^x} = \frac{81}{3^x}$ Now, let $t = 3^x$. Since $3^x$ is always positive for real $x$, we have $t > 0$. Substituting $t$ into the equation: $28 = \frac{t}{3} + \frac{81}{t}$

Step 3: Solve the Quadratic Equation for $t$ To eliminate the fractions, multiply the entire equation by $3t$ (since $t>0$, $3t \neq 0$): $28 \times 3t = \left(\frac{t}{3}\right) \times 3t + \left(\frac{81}{t}\right) \times 3t$ $84t = t^2 + 243$ Rearrange the terms to form a standard quadratic equation: $t^2 - 84t + 243 = 0$ We need to find two numbers that multiply to 243 and add up to -84. By factoring 243, we find that $3 \times 81 = 243$ and $3 + 81 = 84$. Thus, the numbers are -3 and -81. $(t - 3)(t - 81) = 0$ This yields two possible values for $t$: $t = 3$ or $t = 81$.

Step 4: Determine the Valid Value(s) for $\sin 2\alpha$ Recall our substitution $t = 3^x$, where $x = 2\sin 2\alpha$. So, $t = 3^{2\sin 2\alpha}$.

Case 1: $t = 3$ $3^{2\sin 2\alpha} = 3^1$ Equating the exponents: $2\sin 2\alpha = 1$ $\sin 2\alpha = \frac{1}{2}$ This value is valid because $\frac{1}{2}$ is within the range $[-1, 1]$ for the sine function.

Case 2: $t = 81$ $3^{2\sin 2\alpha} = 81$ Since $81 = 3^4$, we have: $3^{2\sin 2\alpha} = 3^4$ Equating the exponents: $2\sin 2\alpha = 4$ $\sin 2\alpha = 2$ This value is invalid because $2$ is outside the range $[-1, 1]$ for the sine function.

Therefore, the only valid value is $\sin 2\alpha = \frac{1}{2}$.

Step 5: Calculate the First Term ($a_1$) and Common Difference ($d$) Now we find the actual value of the first term $a_1$ using $\sin 2\alpha = \frac{1}{2}$: $a_1 = 3^{2\sin 2\alpha - 1}$ $a_1 = 3^{2(\frac{1}{2}) - 1}$ $a_1 = 3^{1 - 1}$ $a_1 = 3^0$ $a_1 = 1$ The second term $a_2$ is given as 14. The common difference $d$ is: $d = a_2 - a_1$ $d = 14 - 1$ $d = 13$

Step 6: Calculate the Sixth Term ($T_6$) The formula for the $n$-th term of an A.P. is $T_n = a_1 + (n-1)d$. We need to find the sixth term ($n=6$): $T_6 = a_1 + (6-1)d$ $T_6 = a_1 + 5d$ Substitute the values $a_1 = 1$ and $d = 13$: $T_6 = 1 + 5 \times 13$ $T_6 = 1 + 65$ $T_6 = 66$

Common Mistakes & Tips

• Trigonometric Range Check: Always verify that the values obtained for trigonometric functions (like $\sin 2\alpha$) lie within their valid range $[-1, 1]$. Any value outside this range must be discarded.
• Substitution Strategy: For problems involving complex exponential or trigonometric expressions, using substitution can greatly simplify the equation, making it easier to solve.
• Careful Algebraic Manipulation: Pay close attention to exponent rules and algebraic steps, especially when solving quadratic equations and simplifying expressions.

Summary The problem requires us to use the property of an arithmetic progression that $2a_2 = a_1 + a_3$. By substituting the given terms and employing a substitution for the exponential part, we transformed the problem into a quadratic equation. After solving for the substitution variable and validating the trigonometric values, we determined the first term and common difference of the A.P. Finally, we used the formula for the $n$-th term of an A.P. to find the sixth term. The sixth term of the A.P. is 66.

The final answer is $\boxed{66}$.