Let the sixth term in the binomial expansion of (√(2^(log₂(10-3ˣ)))) + ⁵√(2^((x-2)log₂3)))ᵐ, in the

Q: Wait, why do we assume the coefficients $C_2, C_3, C_4$ are in A.P. just because they are the 1st, 3rd, and 5th terms of an A.P.?

That is a fantastic observation! If a sequence $x_1, x_2, x_3, x_4, x_5$ is in A.P., then the difference between consecutive terms is a constant $d$. Therefore, $x_3 = x_1 + 2d$ and $x_5 = x_3 + 2d$. This means $x_1, x_3, x_5$ are also in A.P. with a common difference of $2d$. So, yes, the coefficients themselves must satisfy the A.P. condition $2 \cdot C_3 = C_2 + C_4$.

Q: Why did we divide by $m$ in the quadratic equation? Could $m=0$ be a solution?

Great catch! In the equation $m(m-1) = m + \frac{m(m-1)(m-2)}{6}$, we divided by $m$. We can only do this if $m \neq 0$. Since the problem discusses the sixth term of a binomial expansion, $m$ must be at least 5. Thus, $m=0$ is physically impossible in this context. Always check your constraints!

Q: Is it always safe to use $a^{\log_a b} = b$?

It is safe as long as the arguments are valid. Here, we have $\log_2(10-3^x)$ and $\log_2(3^{x-2})$. For these to be defined, $10-3^x > 0$ and $3^{x-2} > 0$. Our solutions $x=0$ and $x=2$ satisfy these conditions perfectly, so the simplification is mathematically sound.

Question

Let the sixth term in the binomial expansion of $(\sqrt{2^{\log_2(10-3^x)}} + \sqrt[5]{2^{(x-2)\log_2 3}})^m$, in the increasing powers of $2^{(x-2)\log_2 3}$, be 21. If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of $x$ is ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of logarithms, binomial coefficients, and exponential functions.

But remember, in the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

Our journey begins with the binomial expression:
$$(\sqrt{2^{\log_2(10-3^x)}} + \sqrt[5]{2^{(x-2)\log_2 3}})^m$$

Look closely at the first term. We have $2^{\log_2(10-3^x)}$. Recall the fundamental identity: $a^{\log_a b} = b$. The base $2$ and the $\log_2$ are essentially inverses, leaving us with $\sqrt{10-3^x}$.

For the second term, we use the property $n \log_a b = \log_a (b^n)$ to rewrite the exponent as $\log_2(3^{x-2})$. Again, the $2$ and $\log_2$ vanish, leaving us with the fifth root of $3^{x-2}$, which is $(3^{x-2})^{1/5}$.

### The Arithmetic Progression Trap

Now, we encounter a classic JEE trap: the binomial coefficients. We are told that the coefficients of the second, third, and fourth terms—let us call them $C_2, C_3, C_4$—are the first, third, and fifth terms of an Arithmetic Progression.

If you have a sequence $a_1, a_2, a_3, a_4, a_5$ in A.P., then $a_1, a_3, a_5$ are also in A.P. with a common difference of $2d$. This implies the condition:
$$2 \cdot C_3 = C_2 + C_4$$

Substituting the standard combination formulas, we get:
$$2 \cdot {}^mC_2 = {}^mC_1 + {}^mC_3$$

Expanding these, we arrive at:
$$2 \cdot \frac{m(m-1)}{2} = m + \frac{m(m-1)(m-2)}{6}$$

Since $m \ge 5$ (because the sixth term exists), we can safely divide by $m$ and solve for $m$. The resulting quadratic equation is:
$$m^2 - 9m + 14 = 0$$

This yields $m=7$ or $m=2$. Since $m$ must be at least 5, we lock in **$m=7$**.

### The Sixth Term Revelation

With $m=7$, we turn to the sixth term, $T_6$. The general term formula is $T_{r+1} = {}^mC_r a^{m-r} b^r$. For $T_6$, we set $r=5$.

Thus, $T_6 = {}^7C_5 a^2 b^5$. We are told this equals 21. Since ${}^7C_5 = {}^7C_2 = 21$, our equation simplifies to:
$$21 \cdot a^2 b^5 = 21 \quad \Rightarrow \quad a^2 b^5 = 1$$

Substituting our simplified $a = (10-3^x)^{1/2}$ and $b = (3^{x-2})^{1/5}$ back in:
$$(10-3^x) \cdot (3^{x-2}) = 1$$

### The Final Quadratic Dance

Let $y = 3^x$. The equation becomes:
$$(10-y) \cdot \frac{y}{9} = 1$$

Multiplying by 9, we get $10y - y^2 = 9$, which rearranges to:
$$y^2 - 10y + 9 = 0$$

Factoring this, we find $(y-9)(y-1) = 0$. This gives us $y=9$ or $y=1$.

If $3^x = 9$, then $x=2$. If $3^x = 1$, then $x=0$. The sum of the squares of these values is:
$$2^2 + 0^2 = 4$$

We have conquered the problem! The final answer is **4**.

Enter Numerical Value:

Visualized Solution

Simplify the First Term $a$

Simplify the Second Term $b$

Set up the A.P. Condition

Expand the Coefficients

Solve for $m$

Set up the Sixth Term $T_{6}$

Substitute $a$ and $b$

Form the Quadratic in $3^{x}$

Solve for $x$

Final Calculation

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

The Arithmetic Progression Trap

The Sixth Term Revelation

The Final Quadratic Dance

Similar Questions

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If the coefficients of $x$ and $x^{2}$ in $(1 + x)^{p} (1 - x)^{q}$ are 4 and $- 5$ respectively, then $2 p + 3 q$ is equal to

Enter Numerical Value:

Visualized Solution

Simplify the First Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Simplify the Second Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Set up the A.P. Condition

Expand the Coefficients

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Set up the Sixth Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T6​

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Form the Quadratic in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3x

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x