If in the expansion of `(a-2b)^(n)`, the sum of `5^(th)` and `6^(th)` terms is 0, then the values of `a/b` is (a)`(n-4)/(5)` (b) `(2(n-4))/(5)` (c) `(5)/(n-4)` (d) `(5)/(2(n-4))`

Options

A. `(n-4)/(5)`
B. `(2(n-4))/(5)`
C. `(5)/(n-4)`
D. `(5)/(2(n-4))`

Explanation

To solve the problem, we need to find the value of \( \frac{a}{b} \) given that the sum of the 5th and 6th terms in the expansion of \( (a - 2b)^n \) is zero.

### Step-by-step Solution:

1. **Identify the General Term**:
The general term \( T_r \) in the expansion of \( (x + y)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} x^{n-r} y^r \]
For our case, \( x = a \) and \( y = -2b \).

2. **Find the 5th and 6th Terms**:
- The 5th term \( T_5 \) corresponds to \( r = 4 \):
\[ T_5 = \binom{n}{4} a^{n-4} (-2b)^4 = \binom{n}{4} a^{n-4} \cdot 16b^4 \]
- The 6th term \( T_6 \) corresponds to \( r = 5 \):
\[ T_6 = \binom{n}{5} a^{n-5} (-2b)^5 = \binom{n}{5} a^{n-5} \cdot (-32b^5) \]

3. **Set Up the Equation**:
According to the problem, the sum of the 5th and 6th terms is zero:
\[ T_5 + T_6 = 0 \]
This implies:
\[ \binom{n}{4} a^{n-4} \cdot 16b^4 + \binom{n}{5} a^{n-5} \cdot (-32b^5) = 0 \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ \binom{n}{4} a^{n-4} \cdot 16b^4 = 32 \cdot \binom{n}{5} a^{n-5} b^5 \]
Dividing both sides by \( a^{n-5} \) and \( b^4 \) (assuming \( a \neq 0 \) and \( b \neq 0 \)):
\[ \binom{n}{4} \cdot 16b = 32 \cdot \binom{n}{5} b \]

5. **Canceling \( b \)**:
Since \( b \neq 0 \), we can divide both sides by \( b \):
\[ \binom{n}{4} \cdot 16 = 32 \cdot \binom{n}{5} \]

6. **Using the Property of Binomial Coefficients**:
We know that:
\[ \binom{n}{5} = \frac{n-4}{5} \cdot \binom{n}{4} \]
Substituting this into the equation:
\[ 16 \cdot \binom{n}{4} = 32 \cdot \frac{n-4}{5} \cdot \binom{n}{4} \]
Canceling \( \binom{n}{4} \) (assuming \( \binom{n}{4} \neq 0 \)):
\[ 16 = \frac{32(n-4)}{5} \]

7. **Solving for \( n \)**:
Multiplying both sides by 5:
\[ 80 = 32(n - 4) \]
Dividing by 32:
\[ n - 4 = \frac{80}{32} = \frac{5}{2} \]
Thus:
\[ n = 4 + \frac{5}{2} = \frac{13}{2} \]

8. **Finding \( \frac{a}{b} \)**:
From our earlier equation, we have:
\[ \frac{a}{b} = \frac{2(n-4)}{5} \]
Substituting \( n = \frac{13}{2} \):
\[ n - 4 = \frac{13}{2} - 4 = \frac{13}{2} - \frac{8}{2} = \frac{5}{2} \]
Therefore:
\[ \frac{a}{b} = \frac{2 \cdot \frac{5}{2}}{5} = 2 \]

### Conclusion:
The value of \( \frac{a}{b} \) is \( 2 \).