Let the coefficients of the middle terms in the expansion of (1/√6 + βx)⁴, (1-3βx)² and (1 - β/2 x)⁶

Q: Wait, why is the middle term $T_{\frac{n}{2}+1}$? I thought it was $T_{\frac{n}{2}}$?

That is a classic confusion! Think about it this way: if you have $n=4$, there are $n+1=5$ terms. The terms are $T_1, T_2, T_3, T_4, T_5$. The middle one is $T_3$. If you used $n/2$, you would get $T_2$, which is wrong. So, the formula $T_{\frac{n}{2}+1}$ is the correct way to land exactly in the center of an odd number of terms.

Q: Can I just use the common difference $d = a_2 - a_1$ and $d = a_3 - a_2$ to solve this?

Absolutely! That is mathematically equivalent to the property $2a_2 = a_1 + a_3$. Using $2a_2 = a_1 + a_3$ is just a faster, more elegant way to skip the intermediate step of calculating $d$ explicitly. Both methods will lead you to the same quadratic equation, so choose the one that feels most intuitive to you.

Q: Why do we reject $\beta = -2$?

Great catch! Always look back at the problem statement. The question explicitly states that $\beta > 0$. In competitive exams like JEE, these constraints are not just filler text; they are crucial for selecting the correct root of your quadratic equation. Always check your final values against the given constraints!

Question

Let the coefficients of the middle terms in the expansion of $\left(\frac{1}{\sqrt{6}} + \beta x\right)^4, (1-3\beta x)^2$ and $\left(1 - \frac{\beta}{2}x\right)^6, \beta > 0$, respectively form the first three terms of an A.P. If $d$ is the common difference of this A.P., then $50 - \frac{2d}{\beta^2}$ is equal to ______.

Sigma Percentile · Accepted Answer

### The Beauty of the Binomial Middle Term

Welcome, future engineer. Today, we are going to dissect a problem that looks like a tangled mess of coefficients and variables, but beneath the surface, it is a beautiful, rhythmic dance of the Binomial Theorem and Arithmetic Progressions.

Take a deep breath. We are going to break this down step-by-step, not just to find the answer, but to understand the logic that governs these expansions.

### Phase 1: Unlocking the Middle Terms

First, let us look at the Binomial Theorem. When we expand $(a+b)^n$, we get $n+1$ terms. If $n$ is even, $n+1$ is odd, which means there is exactly one middle term located at position $T_{\frac{n}{2}+1}$.

For the first expansion, $(\frac{1}{\sqrt{6}} + \beta x)^4$, the power is $n=4$. The middle term is $T_3$. Using the general term formula $T_{r+1} = \binom{n}{r} a^{n-r} b^r$, we set $r=2$:

$$T_3 = \binom{4}{2} \left(\frac{1}{\sqrt{6}}\right)^2 (\beta x)^2 = 6 \cdot \frac{1}{6} \cdot \beta^2 x^2 = \beta^2 x^2$$

Thus, our first coefficient is $a_1 = \beta^2$.

Next, for $(1 - 3\beta x)^2$, the power is $n=2$. The middle term is $T_2$. Setting $r=1$:

$$T_2 = \binom{2}{1} (1)^1 (-3\beta x)^1 = 2 \cdot (-3\beta x) = -6\beta x$$

Here is where many students stumble—do not forget that negative sign! Our second coefficient is $a_2 = -6\beta$.

Finally, for $(1 - \frac{\beta}{2}x)^6$, the power is $n=6$. The middle term is $T_4$. Setting $r=3$:

$$T_4 = \binom{6}{3} (1)^3 \left(-\frac{\beta}{2} x\right)^3 = 20 \cdot \left(-\frac{\beta^3}{8}\right) x^3 = -\frac{5}{2}\beta^3 x^3$$

Our third coefficient is $a_3 = -\frac{5}{2}\beta^3$.

### Phase 2: The Arithmetic Bridge

Now that we have our three coefficients, $a_1 = \beta^2$, $a_2 = -6\beta$, and $a_3 = -\frac{5}{2}\beta^3$, we are told they form an Arithmetic Progression. The defining property of an A.P. is that the difference between consecutive terms is constant, leading to the relation $2a_2 = a_1 + a_3$.

Substituting our values, we get:

$$2(-6\beta) = \beta^2 - \frac{5}{2}\beta^3$$

### Phase 3: The Algebraic Dance

This equation looks a bit intimidating, but let us simplify it. Multiplying by $2$ gives $-24\beta = 2\beta^2 - 5\beta^3$. Since we know $\beta > 0$, we can safely divide by $\beta$:

$$-24 = 2\beta - 5\beta^2 \implies 5\beta^2 - 2\beta - 24 = 0$$

Factoring this quadratic equation, we find:

$$(5\beta - 12)(\beta + 2) = 0$$

Since $\beta$ must be positive, we discard $\beta = -2$ and keep **$\beta = \frac{12}{5}$**.

### Phase 4: The Final Simplification

We need to find $50 - \frac{2d}{\beta^2}$, where $d$ is the common difference. Instead of calculating $d$ immediately, let us express it as $d = a_2 - a_1 = -6\beta - \beta^2$.

Substituting this into our target expression:

$$50 - \frac{2(-6\beta - \beta^2)}{\beta^2} = 50 + \frac{12\beta + 2\beta^2}{\beta^2} = 50 + \frac{12}{\beta} + 2 = 52 + \frac{12}{\beta}$$

Now, plug in $\beta = \frac{12}{5}$:

$$52 + \frac{12}{12/5} = 52 + 5 = 57$$

And there it is! By staying calm and simplifying the expression before plugging in the numbers, we arrived at the elegant solution of **$57$**.

Enter Numerical Value:

Visualized Solution

Middle Term of $(a + b)^{n}$

Middle Term of $(\frac{1}{6} + β x)^{4}$

Middle Term of $(1 - 3 β x)^{2}$

Middle Term of $(1 - \frac{β}{2} x)^{6}$

Applying the A.P. Condition

Simplifying the Equation

Solving for $β$

Defining Common Difference $d$

Simplifying the Target Expression

Final Calculation and Result

The Sigma Insight: General Term and Middle Term

The Beauty of the Binomial Middle Term

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

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Middle Term of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (6​1​+βx)4

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