The coefficient of the middle term in the binomial
expansion , in power of x, of `(1 + ax)^(4) ` and of ` ( 1 - ax)^(6)` is same, a equal .

Options

A. `(3)/(5)`
B. `(10)/(3)`
C. `- (3)/(10)`
D. `- (5)/(3)`

Explanation

To find the value of \( a \) such that the coefficients of the middle terms in the expansions of \( (1 + ax)^4 \) and \( (1 - ax)^6 \) are equal, we will follow these steps:

### Step 1: Identify the middle term in the expansion of \( (1 + ax)^4 \)

The number of terms in the expansion of \( (1 + ax)^4 \) is \( n + 1 = 4 + 1 = 5 \). Since there are 5 terms, the middle term is the 3rd term.

Using the formula for the \( r \)-th term in the binomial expansion, we have:
\[ T_{r+1} = \binom{n}{r} (1)^{n-r} (ax)^r \]
For the 3rd term (\( T_3 \)), \( r = 2 \):
\[ T_3 = \binom{4}{2} (1)^{4-2} (ax)^2 = \binom{4}{2} (ax)^2 \]
Calculating \( \binom{4}{2} \):
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]
Thus, the coefficient of the middle term in \( (1 + ax)^4 \) is:
\[ 6a^2 \]

### Step 2: Identify the middle term in the expansion of \( (1 - ax)^6 \)

The number of terms in the expansion of \( (1 - ax)^6 \) is \( n + 1 = 6 + 1 = 7 \). Since there are 7 terms, the middle term is the 4th term.

Using the same formula for the \( r \)-th term, we have:
\[ T_{r+1} = \binom{n}{r} (1)^{n-r} (-ax)^r \]
For the 4th term (\( T_4 \)), \( r = 3 \):
\[ T_4 = \binom{6}{3} (1)^{6-3} (-ax)^3 = \binom{6}{3} (-ax)^3 \]
Calculating \( \binom{6}{3} \):
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
Thus, the coefficient of the middle term in \( (1 - ax)^6 \) is:
\[ 20(-a^3) = -20a^3 \]

### Step 3: Set the coefficients equal

Since the coefficients of the middle terms are equal, we have:
\[ 6a^2 = -20a^3 \]

### Step 4: Solve for \( a \)

Rearranging gives:
\[ 20a^3 + 6a^2 = 0 \]
Factoring out \( 2a^2 \):
\[ 2a^2(10a + 3) = 0 \]
Setting each factor to zero gives:
1. \( 2a^2 = 0 \) which implies \( a = 0 \)
2. \( 10a + 3 = 0 \) which implies \( a = -\frac{3}{10} \)

### Conclusion

The values of \( a \) that satisfy the equation are \( a = 0 \) or \( a = -\frac{3}{10} \).