The `6^("th")` term of expansion `[sqrt(2^(log_(10)(10-3^(x))))+root(5)(2^((x-2)log_(10)3))]^(m)` is 21 and the coefficient of `2^(nd), 3^(rd) and 4^(th)` terms of it are respectively `1^(st), 3^(rd) and 5^(th)` term of an A.P. Find x.

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem step-by-step, we will break down the given information and apply the necessary mathematical concepts.

### Step 1: Understanding the Problem
We need to find the value of \( x \) given that the 6th term of the expansion

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

is equal to 21, and the coefficients of the 2nd, 3rd, and 4th terms are in arithmetic progression (A.P.).

### Step 2: Identify the Terms in the A.P.
Let the first term of the A.P. be \( a \) and the common difference be \( d \). The terms can be expressed as:
- 1st term: \( a \)
- 2nd term: \( a + d \)
- 3rd term: \( a + 2d \)
- 4th term: \( a + 3d \)
- 5th term: \( a + 4d \)

From the problem, we know:
- Coefficient of the 2nd term = \( a \)
- Coefficient of the 3rd term = \( a + d \)
- Coefficient of the 4th term = \( a + 3d \)

Since these coefficients are in A.P., we have:
\[ 2(a + d) = a + (a + 3d) \]
Simplifying this gives:
\[ 2a + 2d = 2a + 3d \implies 2d = 3d \implies d = 0 \]
This means \( a + d = a \), and all terms are equal.

### Step 3: Calculate the Coefficients
The coefficients of the terms in the binomial expansion can be calculated using the binomial theorem. The \( r \)-th term in the expansion of \( (x + y)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} x^{n-r} y^r \]

For our case:
- 2nd term (\( T_2 \)): \( \binom{m}{1} \left(\sqrt{2^{\log_{10}(10-3^x)}}\right)^{m-1} \left(\sqrt[5]{2^{(x-2)\log_{10}(3)}}\right)^{1} \)
- 3rd term (\( T_3 \)): \( \binom{m}{2} \left(\sqrt{2^{\log_{10}(10-3^x)}}\right)^{m-2} \left(\sqrt[5]{2^{(x-2)\log_{10}(3)}}\right)^{2} \)
- 4th term (\( T_4 \)): \( \binom{m}{3} \left(\sqrt{2^{\log_{10}(10-3^x)}}\right)^{m-3} \left(\sqrt[5]{2^{(x-2)\log_{10}(3)}}\right)^{3} \)

### Step 4: Set Up the Equation for the 6th Term
The 6th term corresponds to \( T_6 \):
\[ T_6 = \binom{m}{5} \left(\sqrt{2^{\log_{10}(10-3^x)}}\right)^{m-5} \left(\sqrt[5]{2^{(x-2)\log_{10}(3)}}\right)^{5} \]
This is given to be equal to 21.

### Step 5: Solve for \( m \)
Using the binomial coefficient:
\[ T_6 = \frac{m!}{5!(m-5)!} \left(\sqrt{2^{\log_{10}(10-3^x)}}\right)^{m-5} \left(2^{(x-2)\log_{10}(3)}\right) \]
Setting this equal to 21 allows us to solve for \( m \).

### Step 6: Solve for \( x \)
Once \( m \) is found, we can substitute back into the equations for the coefficients to find \( x \).

### Final Steps: Calculate and Verify
After substituting the values and simplifying, we will find \( x \).