If a > 0 and the coefficient of `x^(5)` in the expansion of `(1+ax)^(2)(1-x)^(7)` is `-21` then a=________

Answer

To solve the problem, we need to find the value of \( a \) such that the coefficient of \( x^5 \) in the expansion of \( (1 + ax)^2 (1 - x)^7 \) is equal to \(-21\).

### Step-by-Step Solution:

1. **Expand the Expression**:
We start with the expression \( (1 + ax)^2 (1 - x)^7 \).

2. **Find the Coefficient of \( x^5 \)**:
We can find the coefficient of \( x^5 \) by considering the contributions from both parts of the product:
- From \( (1 + ax)^2 \), we can get terms \( 1 \), \( 2ax \), and \( a^2x^2 \).
- From \( (1 - x)^7 \), we can use the binomial expansion.

3. **Use the Binomial Theorem**:
The binomial expansion of \( (1 - x)^7 \) gives us:
\[ (1 - x)^7 = \sum_{r=0}^{7} \binom{7}{r} (-1)^r x^r \]
We need to find the terms that contribute to \( x^5 \).

4. **Identify Relevant Terms**:
We can get \( x^5 \) from:
- \( 1 \) from \( (1 + ax)^2 \) and \( x^5 \) from \( (1 - x)^7 \).
- \( 2ax \) from \( (1 + ax)^2 \) and \( x^4 \) from \( (1 - x)^7 \).
- \( a^2x^2 \) from \( (1 + ax)^2 \) and \( x^3 \) from \( (1 - x)^7 \).

5. **Calculate Each Contribution**:
- The coefficient of \( x^5 \) from \( 1 \cdot x^5 \) is \( \binom{7}{5} (-1)^5 = -21 \).
- The coefficient of \( x^4 \) from \( 2ax \) is \( 2a \cdot \binom{7}{4} (-1)^4 = 2a \cdot 35 = 70a \).
- The coefficient of \( x^3 \) from \( a^2x^2 \) is \( a^2 \cdot \binom{7}{3} (-1)^3 = a^2 \cdot (-35) = -35a^2 \).

6. **Set Up the Equation**:
Now, we combine these contributions to find the total coefficient of \( x^5 \):
\[ -21 + 70a - 35a^2 = -21 \]
Simplifying gives:
\[ 70a - 35a^2 = 0 \]

7. **Factor the Equation**:
Factor out \( 35a \):
\[ 35a(2 - a) = 0 \]

8. **Solve for \( a \)**:
Setting each factor to zero gives:
- \( 35a = 0 \) which implies \( a = 0 \) (not valid since \( a > 0 \)).
- \( 2 - a = 0 \) which implies \( a = 2 \).

9. **Conclusion**:
Since \( a > 0 \), the only valid solution is:
\[ a = 2 \]

### Final Answer:
The value of \( a \) is \( \boxed{2} \).