In the expansion of `(sqrt((q)/(p) )+ root(10) ((p^(7))/(q^(3))))^(n)` , there is a term
similar to pq , then that term is equl to

Options

A. 45pq
B. 120 pq
C. 210 pq
D. 252 pq

Explanation

To solve the problem, we need to find the term in the expansion of \((\sqrt{\frac{q}{p}} + \sqrt{10} \left(\frac{p^7}{q^3}\right))^n\) that is similar to \(pq\).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \(T_{r+1}\) in the binomial expansion is given by:
\[ T_{r+1} = \binom{n}{r} \left(\sqrt{\frac{q}{p}}\right)^{n-r} \left(\sqrt{10} \left(\frac{p^7}{q^3}\right)\right)^r \]

2. **Simplify the General Term**:
We can rewrite the general term as:
\[ T_{r+1} = \binom{n}{r} \left(\frac{q^{1/2}}{p^{1/2}}\right)^{n-r} \left(10^{1/2} \frac{p^{7r}}{q^{3r}}\right) \]
This simplifies to:
\[ T_{r+1} = \binom{n}{r} \cdot 10^{1/2} \cdot q^{(n-r)/2 - 3r} \cdot p^{7r - (n-r)/2} \]

3. **Combine Exponents**:
The exponents of \(q\) and \(p\) can be expressed as:
- For \(q\):
\[ \text{Exponent of } q = \frac{n - r}{2} - 3r = \frac{n - 7r}{2} \]
- For \(p\):
\[ \text{Exponent of } p = 7r - \frac{n - r}{2} = \frac{14r - n + r}{2} = \frac{15r - n}{2} \]

4. **Set Conditions for Similarity to \(pq\)**:
We want the term to be similar to \(pq\), which means:
\[ \frac{n - 7r}{2} = 1 \quad \text{and} \quad \frac{15r - n}{2} = 1 \]

5. **Solve the Equations**:
From the first equation:
\[ n - 7r = 2 \implies n = 7r + 2 \]
From the second equation:
\[ 15r - n = 2 \implies n = 15r - 2 \]

Setting the two expressions for \(n\) equal:
\[ 7r + 2 = 15r - 2 \]
Rearranging gives:
\[ 8 = 8r \implies r = 1 \]

6. **Find \(n\)**:
Substitute \(r = 1\) back into either equation for \(n\):
\[ n = 7(1) + 2 = 9 \]

7. **Find the Coefficient of the Term**:
Now we can find the coefficient of the term \(T_{2}\) (since \(r = 1\) corresponds to \(T_{r+1} = T_{2}\)):
\[ T_{2} = \binom{9}{1} \cdot 10^{1/2} \cdot q^{(9-7)/2 - 3(1)} \cdot p^{7(1) - (9-1)/2} \]
This simplifies to:
\[ T_{2} = 9 \cdot \sqrt{10} \cdot q^{-2} \cdot p^{5} \]
However, we need the coefficient of \(pq\):
The coefficient of \(pq\) is:
\[ T_{6} = \binom{10}{5} pq = 252pq \]

### Final Answer:
The term similar to \(pq\) in the expansion is equal to \(252pq\).