Write down and simplify
`10^("th")" term in "((3p)/(4)-5q)^(14)`

Answer

To find the 10th term in the binomial expansion of \(\left(\frac{3p}{4} - 5q\right)^{14}\), we will use the Binomial Theorem, which states that the \(r+1\)th term in the expansion of \((a + b)^n\) is given by:

\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

### Step-by-Step Solution:

1. **Identify the components of the binomial expansion**:
- Here, \(a = \frac{3p}{4}\), \(b = -5q\), and \(n = 14\).

2. **Determine the term number**:
- We need to find the 10th term, which corresponds to \(r = 9\) (since \(T_{r+1}\) means \(r+1\)).

3. **Apply the Binomial Theorem**:
- The 10th term \(T_{10}\) can be expressed as:
\[ T_{10} = \binom{14}{9} \left(\frac{3p}{4}\right)^{14-9} (-5q)^9 \]

4. **Simplify the expression**:
- Calculate \(\binom{14}{9}\):
\[ \binom{14}{9} = \binom{14}{5} = \frac{14!}{9! \cdot 5!} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002 \]
- Now substitute back into the term:
\[ T_{10} = 2002 \left(\frac{3p}{4}\right)^5 (-5q)^9 \]

5. **Calculate \(\left(\frac{3p}{4}\right)^5\) and \((-5q)^9\)**:
- \(\left(\frac{3p}{4}\right)^5 = \frac{(3^5)(p^5)}{4^5} = \frac{243p^5}{1024}\)
- \((-5q)^9 = -5^9 q^9 = -1953125 q^9\)

6. **Combine the results**:
\[ T_{10} = 2002 \cdot \frac{243p^5}{1024} \cdot (-1953125 q^9) \]

7. **Final simplification**:
- Combine the constants:
\[ T_{10} = -2002 \cdot 243 \cdot 1953125 \cdot \frac{p^5 q^9}{1024} \]

### Final Answer:
\[ T_{10} = -\frac{2002 \cdot 243 \cdot 1953125}{1024} p^5 q^9 \]