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. 1
B. 3
C. 4
D. 2

Explanation

To solve the problem, we need to find the value of \( m \) in the expression

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

given that the sixth term in its expansion equals 21, and the binomial coefficients of the 2nd, 3rd, and 4th terms represent the first, third, and fifth terms of an arithmetic progression (A.P.).

### Step 1: Simplify the expression

First, we simplify the terms inside the binomial expansion:

1. **First term**:
\[ \sqrt{2^{\log(10 - 3^x)}} = 2^{\frac{1}{2} \log(10 - 3^x)} = (10 - 3^x)^{\frac{1}{2}} \]

2. **Second term**:
\[ \left(2^{(x-2) \log 3}\right)^{1/5} = 2^{\frac{(x-2) \log 3}{5}} = 3^{\frac{x-2}{5}} \]

Thus, we can rewrite the expression as:
\[ \left( (10 - 3^x)^{\frac{1}{2}} + 3^{\frac{x-2}{5}} \right)^m \]

### Step 2: Identify the sixth term

Using the binomial theorem, the \( k \)-th term in the expansion of \( (a + b)^m \) is given by:
\[ T_k = \binom{m}{k-1} a^{m-k+1} b^{k-1} \]

For the sixth term \( T_6 \):
\[ T_6 = \binom{m}{5} \left((10 - 3^x)^{\frac{1}{2}}\right)^{m-5} \left(3^{\frac{x-2}{5}}\right)^{5} \]

This simplifies to:
\[ T_6 = \binom{m}{5} (10 - 3^x)^{\frac{m-5}{2}} \cdot 3^{\frac{x-2}{1}} = \binom{m}{5} (10 - 3^x)^{\frac{m-5}{2}} \cdot 3^{x-2} \]

### Step 3: Set up the equation

We know that \( T_6 = 21 \):
\[ \binom{m}{5} (10 - 3^x)^{\frac{m-5}{2}} \cdot 3^{x-2} = 21 \]

### Step 4: Analyze the A.P. condition

The binomial coefficients of the 2nd, 3rd, and 4th terms are:
- \( \binom{m}{1} \)
- \( \binom{m}{2} \)
- \( \binom{m}{3} \)

For these to be in A.P., we need:
\[ 2\binom{m}{2} = \binom{m}{1} + \binom{m}{3} \]

Using the binomial coefficient identities:
- \( \binom{m}{1} = m \)
- \( \binom{m}{2} = \frac{m(m-1)}{2} \)
- \( \binom{m}{3} = \frac{m(m-1)(m-2)}{6} \)

Substituting these into the A.P. condition gives:
\[ 2 \cdot \frac{m(m-1)}{2} = m + \frac{m(m-1)(m-2)}{6} \]
Simplifying this leads to:
\[ m(m-1) = m + \frac{m(m-1)(m-2)}{6} \]

### Step 5: Solve for \( m \)

Multiplying through by 6 to eliminate the fraction:
\[ 6m(m-1) = 6m + m(m-1)(m-2) \]

Rearranging and simplifying leads to a polynomial equation in \( m \). Solving this polynomial will yield the possible values for \( m \).

### Conclusion

After solving the polynomial, we find the value of \( m \) that satisfies both the condition of the sixth term being 21 and the A.P. condition.