The exponent of x occuring in the 7th term of the expansion of `((ax)/2-8/(bx))^(9)` is

Options

A. 3
B. -3
C. 5
D. -5

Answer

To find the exponent of \( x \) occurring in the 7th term of the expansion of \( \left( \frac{ax}{2} - \frac{8}{bx} \right)^{9} \), we can follow these steps:

### Step 1: Identify the General Term
The general term \( T_{r+1} \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \( n = 9 \), \( a = \frac{ax}{2} \), and \( b = -\frac{8}{bx} \).

### Step 2: Find the 7th Term
To find the 7th term, we need to calculate \( T_7 \), which corresponds to \( r = 6 \) (since \( T_{r+1} \) means we take \( r \) as one less than the term number):
\[ T_7 = \binom{9}{6} \left( \frac{ax}{2} \right)^{9-6} \left( -\frac{8}{bx} \right)^6 \]

### Step 3: Calculate the Binomial Coefficient
Calculate \( \binom{9}{6} \):
\[ \binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]

### Step 4: Substitute Values into the Term
Now substitute the values into the term:
\[ T_7 = 84 \left( \frac{ax}{2} \right)^{3} \left( -\frac{8}{bx} \right)^{6} \]

### Step 5: Simplify the Expression
Calculate \( \left( \frac{ax}{2} \right)^{3} \) and \( \left( -\frac{8}{bx} \right)^{6} \):
\[ \left( \frac{ax}{2} \right)^{3} = \frac{a^3 x^3}{2^3} = \frac{a^3 x^3}{8} \]
\[ \left( -\frac{8}{bx} \right)^{6} = \frac{(-8)^6}{(bx)^6} = \frac{262144}{b^6 x^6} \]

### Step 6: Combine the Terms
Now combine these results:
\[ T_7 = 84 \cdot \frac{a^3 x^3}{8} \cdot \frac{262144}{b^6 x^6} \]
\[ = \frac{84 \cdot 262144 \cdot a^3}{8 \cdot b^6} \cdot \frac{x^3}{x^6} \]
\[ = \frac{84 \cdot 262144 \cdot a^3}{8 \cdot b^6} \cdot x^{-3} \]

### Step 7: Determine the Exponent of \( x \)
From the expression \( x^{-3} \), we can see that the exponent of \( x \) is:
\[ -3 \]

### Final Answer
Thus, the exponent of \( x \) occurring in the 7th term of the expansion is \( -3 \).

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