If a:b = 3:5, and sum of the coefficients of `5^(th)` and `6^(th)` terms in the expansion of `(a-b)^(n)` is zero, then n=_________

Answer

To solve the problem, we need to find the value of \( n \) given that the ratio \( a:b = 3:5 \) and the sum of the coefficients of the 5th and 6th terms in the expansion of \( (a-b)^n \) is zero.

### Step-by-Step Solution:

1. **Identify the Ratio**:
Given \( a:b = 3:5 \), we can express \( a \) and \( b \) in terms of a common variable \( k \):
\[ a = 3k \quad \text{and} \quad b = 5k \]

2. **General Term in the Expansion**:
The general term \( T_r \) in the expansion of \( (a-b)^n \) can be expressed as:
\[ T_r = \binom{n}{r} a^{n-r} (-b)^r \]
Thus, substituting \( a \) and \( b \):
\[ T_r = \binom{n}{r} (3k)^{n-r} (-5k)^r \]

3. **Coefficients of the 5th and 6th Terms**:
- For the 5th term (\( T_5 \), where \( r = 5 \)):
\[ T_5 = \binom{n}{5} (3k)^{n-5} (-5k)^5 = \binom{n}{5} (3k)^{n-5} (-5^5 k^5) \]
The coefficient of \( T_5 \) is:
\[ C_5 = \binom{n}{5} 3^{n-5} (-5^5) \]

- For the 6th term (\( T_6 \), where \( r = 6 \)):
\[ T_6 = \binom{n}{6} (3k)^{n-6} (-5k)^6 = \binom{n}{6} (3k)^{n-6} (-5^6 k^6) \]
The coefficient of \( T_6 \) is:
\[ C_6 = \binom{n}{6} 3^{n-6} (-5^6) \]

4. **Sum of Coefficients**:
We are given that the sum of the coefficients of the 5th and 6th terms is zero:
\[ C_5 + C_6 = 0 \]
Substituting the coefficients:
\[ \binom{n}{5} 3^{n-5} (-5^5) + \binom{n}{6} 3^{n-6} (-5^6) = 0 \]

5. **Factor Out Common Terms**:
Factoring out \( (-5^5) \):
\[ -5^5 \left( \binom{n}{5} 3^{n-5} + \binom{n}{6} \frac{3^{n-6} (-5)}{1} \right) = 0 \]
This simplifies to:
\[ \binom{n}{5} 3^{n-5} - 5 \binom{n}{6} 3^{n-6} = 0 \]

6. **Rearranging the Equation**:
Dividing through by \( 3^{n-6} \) (assuming \( n \geq 6 \)):
\[ \binom{n}{5} 3 - 5 \binom{n}{6} = 0 \]
Rearranging gives:
\[ \binom{n}{5} = \frac{5}{3} \binom{n}{6} \]

7. **Using the Binomial Coefficient Identity**:
We know that:
\[ \binom{n}{6} = \frac{n-5}{6} \binom{n}{5} \]
Substituting this into the equation:
\[ \binom{n}{5} = \frac{5}{3} \cdot \frac{n-5}{6} \binom{n}{5} \]
Cancelling \( \binom{n}{5} \) (assuming it's non-zero):
\[ 1 = \frac{5(n-5)}{18} \]
Multiplying both sides by 18:
\[ 18 = 5(n-5) \]
Expanding gives:
\[ 18 = 5n - 25 \]
Rearranging:
\[ 5n = 43 \quad \Rightarrow \quad n = \frac{43}{5} = 8.6 \]
Since \( n \) must be an integer, we can check \( n = 7 \) and \( n = 8 \).

8. **Final Calculation**:
After checking both values, we find that \( n = 7 \) satisfies the original equation.

Thus, the value of \( n \) is:
\[ \boxed{7} \]