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. 64
B. 32
C. 128
D. none of these

Explanation

To solve the 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, given that the binomial coefficients of the 2nd, 3rd, and 4th terms form an arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Identify the expression**:
The expression can be simplified as follows:
\[ \sqrt{2^{\log(10 - 3^x)}} = 2^{\frac{1}{2} \log(10 - 3^x)} = (10 - 3^x)^{\frac{1}{2}} \]
and
\[ \left(2^{(x-2)\log 3}\right)^{\frac{1}{5}} = 2^{\frac{(x-2)\log 3}{5}}. \]
Thus, we can rewrite the expression as:
\[ \left((10 - 3^x)^{\frac{1}{2}} + 2^{\frac{(x-2)\log 3}{5}}\right)^m. \]

2. **Write the general term of the binomial expansion**:
The general term \( T_k \) in the binomial expansion is given by:
\[ T_k = \binom{m}{k} a^{m-k} b^k, \]
where \( a = (10 - 3^x)^{\frac{1}{2}} \) and \( b = 2^{\frac{(x-2)\log 3}{5}} \).

3. **Find the sixth term**:
The sixth term corresponds to \( k = 5 \):
\[ T_5 = \binom{m}{5} (10 - 3^x)^{\frac{m}{2}} \left(2^{\frac{(x-2)\log 3}{5}}\right)^5. \]
Simplifying this gives:
\[ T_5 = \binom{m}{5} (10 - 3^x)^{\frac{m}{2}} \cdot 2^{(x-2)\log 3}. \]

4. **Set the sixth term equal to 21**:
We know that:
\[ T_5 = 21. \]
Therefore, we have:
\[ \binom{m}{5} (10 - 3^x)^{\frac{m}{2}} \cdot 2^{(x-2)\log 3} = 21. \]

5. **Use the condition of A.P.**:
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 must have:
\[ 2 \binom{m}{2} = \binom{m}{1} + \binom{m}{3}. \]
Expanding this gives:
\[ 2 \cdot \frac{m(m-1)}{2} = m + \frac{m(m-1)(m-2)}{6}. \]
Simplifying leads to:
\[ m(m-1) = 6m + m(m-1)(m-2)/6. \]

6. **Rearranging the equation**:
After simplification, we get:
\[ 6m(m-1) = 6m + m^3 - 3m^2 + 2m. \]
Rearranging gives:
\[ m^3 - 9m + 14 = 0. \]

7. **Finding the roots**:
We can factor this polynomial:
\[ (m - 2)(m - 7)(m + 1) = 0. \]
Thus, \( m = 2, 7, -1 \).

8. **Determine the valid value of \( m \)**:
Since \( m \) must be a positive integer and should allow for the existence of the 3rd term, we discard \( m = 2 \). Hence, the only valid solution is:
\[ m = 7. \]

### Conclusion:
The value of \( m \) is \( 7 \).