If 3^(2 sin 2α - 1), 14 and 3^(4 - 2 sin 2α) are the first three terms of an A.P. for some α, then t

Q: Wait, why do we reject the root $t=81$? Is it mathematically wrong?

It is not algebraically wrong, but it is physically impossible given the constraints of the problem. We defined $t = 3^{2 \sin 2\alpha}$. If $t=81$, then $3^{2 \sin 2\alpha} = 3^4$, implying $2 \sin 2\alpha = 4$, or $\sin 2\alpha = 2$. Since the range of the sine function is $[-1, 1]$, $\sin 2\alpha$ can never be $2$. Always check your domain constraints!

Q: Can I solve for $\alpha$ directly instead of using a substitution variable?

Technically, yes, but it is a trap. Solving for $\alpha$ directly involves solving transcendental equations which are prone to calculation errors. Using a substitution variable like $t$ simplifies the problem into a standard quadratic, which is much safer and faster in a high-pressure exam environment.

Q: Why did we use $2B = A + C$ instead of finding the common difference $d$ first?

We didn't know the common difference $d$ initially! The property $2B = A + C$ is the most direct way to link the three terms without needing to know $d$ or the first term $a$ beforehand. It is the most efficient path to finding the unknown variable $\alpha$.

Question

If $3^{2 \sin 2\alpha - 1}, 14$ and $3^{4 - 2 \sin 2\alpha}$ are the first three terms of an A.P. for some $\alpha$, then the sixth term of this A.P is :

Options:

(A) 65

(B) 78

(C) 81

(D) 66

Sigma Percentile · Accepted Answer

Welcome, future engineers and mathematicians. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of exponents and trigonometry.

You see an expression like $3^{2 \sin 2\alpha - 1}$ and your brain might instinctively want to run away. But I want you to pause. Take a deep breath.

In the world of JEE Advanced, intimidation is the first trap. Let us look past the scary notation and find the elegant structure hidden underneath.

### The Logic Bridge: Unlocking the A.P. Property

The problem tells us that $3^{2 \sin 2\alpha - 1}$, $14$, and $3^{4 - 2 \sin 2\alpha}$ are the first three terms of an Arithmetic Progression (A.P.). What is the soul of an A.P.? It is the constant difference between terms.

If we have three terms $A, B, C$, then $B - A = C - B$, which simplifies beautifully to $2B = A + C$. This is our logic bridge. It allows us to ignore the complexity of the terms for a moment and focus on their relationship.

By substituting our values, we get:

$$2(14) = 3^{2 \sin 2\alpha - 1} + 3^{4 - 2 \sin 2\alpha}$$

This simplifies to:

$$28 = 3^{2 \sin 2\alpha - 1} + 3^{4 - 2 \sin 2\alpha}$$

We have successfully translated a sequence problem into an algebraic equation.

### The Exponent Jungle: Simplifying the Chaos

Now, we face the exponents. They look messy, but they follow the fundamental laws of algebra. Recall that $a^{m-n} = \frac{a^m}{a^n}$.

Applying this, our equation becomes:

$$28 = \frac{3^{2 \sin 2\alpha}}{3^1} + \frac{3^4}{3^{2 \sin 2\alpha}}$$

Suddenly, the fog clears. Do you see it? The term $3^{2 \sin 2\alpha}$ is repeating.

This is the moment where a student becomes an engineer. We don't fight the complexity; we simplify it. Let us introduce a dummy variable, $t = 3^{2 \sin 2\alpha}$.

Our equation transforms into:

$$28 = \frac{t}{3} + \frac{81}{t}$$

This is no longer a trigonometric nightmare; it is a simple rational equation.

### The Quadratic Transformation

To solve for $t$, we clear the denominators by multiplying the entire equation by $3t$. This gives us $28(3t) = t^2 + 243$, which rearranges into the standard quadratic form:

$$t^2 - 84t + 243 = 0$$

Now, we factorize. We need two numbers that multiply to $243$ and add to $-84$. A quick mental check reveals $81$ and $3$.

Thus, $(t - 81)(t - 3) = 0$. We have two candidates: $t = 81$ and $t = 3$.

### The Reality Check: The Sine Constraint

Here is where the JEE examiners test your maturity. We have two values for $t$, but are they both physically possible?

Recall that $t = 3^{2 \sin 2\alpha}$. If $t = 81$, then $3^{2 \sin 2\alpha} = 3^4$, implying $2 \sin 2\alpha = 4$, or $\sin 2\alpha = 2$.

But wait! The range of the sine function is strictly $[-1, 1]$. It can never be $2$. This solution is an illusion, a mathematical ghost. We must reject it.

However, if $t = 3$, then $3^{2 \sin 2\alpha} = 3^1$, implying $2 \sin 2\alpha = 1$, or $\sin 2\alpha = \frac{1}{2}$. This is perfectly valid.

### The Final Victory

With $\sin 2\alpha = \frac{1}{2}$, we can finally find the first term $a$. The expression for the first term was $3^{2 \sin 2\alpha - 1}$.

Substituting our value, we get:

$$a = 3^{1 - 1} = 3^0 = 1$$

Now that we have the first term $a = 1$ and the second term $14$, the common difference $d$ is simply $14 - 1 = 13$.

To find the sixth term, we use the general formula $T_n = a + (n - 1)d$. For $n = 6$, we have:

$$T_6 = 1 + (6 - 1)13 = 1 + 5(13) = 1 + 65 = 66$$

And there it is. Through logic, substitution, and a healthy dose of skepticism regarding our constraints, we have arrived at the answer: **66**.

Never fear the complexity of a problem; just break it down, one step at a time.

Animated Solution for Mathematics - Sequence and Series: If $3^{2 s i n 2 α - 1}, 14$ and $3^{4 - 2 s i n 2 α}$ are the first three terms of an A.P. for some $α$ , then the sixth term of this A.P is :

Select Answer:

Visualized Solution

Given Terms in A.P.

Condition for A.P.

Applying the A.P. Condition

Using Exponent Rules

Substitution for Simplification

Forming the Quadratic Equation

Solving the Quadratic Equation

Checking Constraints for $t$

Finding the First Term $a$

Finding the Common Difference $d$

Calculating the 6th Term

The Sigma Insight: Arithmetic Progression (A.P.)

The Logic Bridge

The Exponent Jungle

The Quadratic Transformation

The Reality Check

The Final Victory

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