Let a and b be two nonzero real numbers. If the coefficient of x⁵ in the expansion of (ax² + 70/(27b

Q: Why do we use $T_{r+1}$ instead of expanding the whole binomial?

Expanding $(ax^2 + \frac{70}{27bx})^4$ is manageable, but $(ax - \frac{1}{bx^2})^7$ would be a nightmare! The general term formula is the surgical tool that lets us extract exactly what we need without the collateral damage of calculating every term.

Q: Can I just cancel $a^3$ without checking if $a=0$?

Never! The problem explicitly states $a$ is a nonzero real number. If you cancel a variable without knowing it's non-zero, you might lose a solution. Always check the constraints.

Q: What happens to the negative sign in the second expansion?

Great catch! In $(ax - \frac{1}{bx^2})^7$, the term is $(-\frac{1}{bx^2})^k$. Since $k=4$ (an even number), the negative sign becomes positive. If $k$ had been odd, that negative sign would have flipped the entire coefficient!

Question

Let $a$ and $b$ be two nonzero real numbers. If the coefficient of $x^5$ in the expansion of $(ax^2 + \frac{70}{27bx})^4$ is equal to the coefficient of $x^{-5}$ in the expansion of $(ax - \frac{1}{bx^2})^7$, then the value of $2b$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the road to JEE mastery! Today, we are not just solving a problem; we are conducting a symphony. The Binomial Theorem is one of the most elegant structures in algebra, and when we are asked to find coefficients in expansions, we are essentially looking for a specific note in a complex melody.

Let us break down this problem, step by step, with the precision of a mathematician and the heart of a student.

### Phase 1: The First Expansion

We start with the expression $(ax^2 + \frac{70}{27bx})^4$. Our goal is to find the coefficient of $x^5$.

Instead of expanding this entire polynomial, we use our surgical tool: the general term formula, $T_{r+1} = \binom{n}{r} X^{n-r} Y^r$. Here, $n=4$, $X=ax^2$, and $Y=\frac{70}{27bx}$.

When we plug these into our formula, we get:

$$T_{r+1} = \binom{4}{r} (ax^2)^{4-r} \left( \frac{70}{27bx} \right)^r$$

Now, here is the secret: do not let the variables and constants mingle. Separate them immediately! By isolating the constants, we get:

$$T_{r+1} = \binom{4}{r} a^{4-r} \left( \frac{70}{27b} \right)^r (x^2)^{4-r} (x^{-1})^r$$

Simplifying the powers of $x$, we find $x^{8-2r} \cdot x^{-r} = x^{8-3r}$. To find the coefficient of $x^5$, we set $8-3r = 5$, which gives us $r=1$.

Substituting $r=1$ back into our constant part, we get:

$$\binom{4}{1} a^3 \left( \frac{70}{27b} \right) = \frac{280a^3}{27b}$$

Keep this value close; it is our first anchor.

### Phase 2: The Second Expansion

Now, we turn our attention to the second expansion: $(ax - \frac{1}{bx^2})^7$. The process is identical, but the stakes are higher.

We use $T_{k+1} = \binom{7}{k} (ax)^{7-k} (-\frac{1}{bx^2})^k$. Again, separate the constants:

$$T_{k+1} = \binom{7}{k} a^{7-k} (-1)^k \left( \frac{1}{b} \right)^k x^{7-k} (x^{-2})^k$$

The power of $x$ becomes $x^{7-k-2k} = x^{7-3k}$. We need the coefficient of $x^{-5}$, so we set $7-3k = -5$, which leads to $3k = 12$, or $k=4$.

Substituting $k=4$ into our constant part, we get:

$$\binom{7}{4} a^3 (-1)^4 \left( \frac{1}{b^4} \right) = 35 a^3 \left( \frac{1}{b^4} \right) = \frac{35a^3}{b^4}$$

### Phase 3: The Grand Equivalence

We have arrived at the climax of our journey. The problem states that these two coefficients are equal. So, we set them against each other:

$$\frac{280a^3}{27b} = \frac{35a^3}{b^4}$$

Because $a$ is a nonzero real number, we can confidently divide both sides by $a^3$. This leaves us with:

$$\frac{280}{27b} = \frac{35}{b^4}$$

Now, watch the magic of algebra. Cross-multiplying gives us:

$$\frac{b^4}{b} = \frac{35 \cdot 27}{280}$$

This simplifies to $b^3 = \frac{35 \cdot 27}{280}$. Since $280 = 35 \cdot 8$, the $35$ cancels out, leaving $b^3 = \frac{27}{8}$.

Taking the cube root, we find $b = \frac{3}{2}$. The question asks for $2b$, so $2 \cdot (\frac{3}{2}) = 3$.

### Conclusion

Wasn't that beautiful? We didn't need to expand the entire binomial. We used the structure of the general term to zoom in on exactly what we needed.

Remember, in JEE, the math is not just about calculation; it is about finding the most efficient path through the forest. Keep practicing, keep visualizing, and most importantly, keep falling in love with the logic! The final answer is **3**.

Enter Numerical Value:

Visualized Solution

Problem Overview & Strategy

General Term for First Expansion

Simplifying Powers of $x$

Coefficient of $x^{5}$

General Term for Second Expansion

Simplifying Powers of $x$ (Second Expansion)

Coefficient of $x^{- 5}$

Equating the Coefficients

Solving for $b$ and Final Answer

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Conclusion

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Enter Numerical Value:

Visualized Solution

Problem Overview & Strategy

General Term for First Expansion

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Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x5

General Term for Second Expansion

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Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x−5