The sixth term in the expansion of `( sqrt(2^(log(10-3^x))) + (2^((x-2)log3))^(1/5))^m` is equal to 21, if it is known that the binomial coefficient of the 2nd 3rd and 4th terms in the expansion represent, respectively, the first, third and fifth terms of an A.P. (the symbol log stands for logarithm to the base 10) The value of m is

Options

A. 6
B. 7
C. 8
D. 9

Explanation

To solve the given problem, we need to find the value of \( m \) such that the sixth term in the expansion of

\[ \left( \sqrt{2^{\log(10 - 3^x)}} + \left(2^{(x-2)\log 3}\right)^{\frac{1}{5}}\right)^m \]

is equal to 21, and the binomial coefficients of the second, third, and fourth terms represent the first, third, and fifth terms of an arithmetic progression (A.P.).

### Step 1: Understanding the Binomial Expansion
The general term \( T_k \) in the binomial expansion of \( (a + b)^m \) is given by:

\[ T_k = \binom{m}{k-1} a^{m - (k-1)} b^{k-1} \]

For our case, \( a = \sqrt{2^{\log(10 - 3^x)}} \) and \( b = \left(2^{(x-2)\log 3}\right)^{\frac{1}{5}} \).

### Step 2: Finding the Sixth Term
The sixth term \( T_6 \) corresponds to \( k = 6 \):

\[ T_6 = \binom{m}{5} a^{m-5} b^5 \]

### Step 3: Finding the Coefficients for A.P.
The binomial coefficients of the second, third, and fourth terms are \( \binom{m}{1} \), \( \binom{m}{2} \), and \( \binom{m}{3} \). For these coefficients to be in A.P., we must have:

\[ 2\binom{m}{2} = \binom{m}{1} + \binom{m}{3} \]

### Step 4: Expanding the Binomial Coefficients
Using the formulas for binomial coefficients:

- \( \binom{m}{1} = m \)
- \( \binom{m}{2} = \frac{m(m-1)}{2} \)
- \( \binom{m}{3} = \frac{m(m-1)(m-2)}{6} \)

Substituting these into the A.P. condition gives:

\[ 2 \cdot \frac{m(m-1)}{2} = m + \frac{m(m-1)(m-2)}{6} \]

### Step 5: Simplifying the Equation
This simplifies to:

\[ m(m-1) = m + \frac{m(m-1)(m-2)}{6} \]

Multiplying through by 6 to eliminate the fraction:

\[ 6m(m-1) = 6m + m(m-1)(m-2) \]

### Step 6: Rearranging the Equation
Rearranging gives:

\[ 6m^2 - 6m = 6m + m^3 - 3m^2 + 2m \]

Combining like terms results in:

\[ m^3 - 3m^2 + 2m - 6m + 6m^2 = 0 \]

This simplifies to:

\[ m^3 + 3m^2 - 4m = 0 \]

Factoring out \( m \):

\[ m(m^2 + 3m - 4) = 0 \]

### Step 7: Solving the Quadratic Equation
The quadratic \( m^2 + 3m - 4 = 0 \) can be solved using the quadratic formula:

\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2} \]

This gives:

\[ m = 1 \quad \text{or} \quad m = -4 \]

### Step 8: Considering Valid Values of \( m \)
Since \( m \) must be a positive integer, we discard \( m = -4 \).

### Conclusion
Thus, the only valid solution for \( m \) is:

\[ \boxed{2} \]