Write down the fourth term in the binomial expansion of `(px + (1)/(x) )^(n)`. If this term is independent of `x`, find the value of n. With this value of `n`, calculate the value of `p` given that the fourth term is equal to `(5)/(2)`.

Answer

To solve the problem, we need to find the fourth term in the binomial expansion of \((px + \frac{1}{x})^n\), determine the value of \(n\) for which this term is independent of \(x\), and subsequently find the value of \(p\) given that the fourth term equals \(\frac{5}{2}\).

### Step 1: Write the General Term
The general term \(T_{r+1}\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = px\) and \(b = \frac{1}{x}\). Thus, the general term becomes:
\[ T_{r+1} = \binom{n}{r} (px)^{n-r} \left(\frac{1}{x}\right)^r \]

### Step 2: Write the Fourth Term
The fourth term \(T_4\) corresponds to \(r = 3\):
\[ T_4 = \binom{n}{3} (px)^{n-3} \left(\frac{1}{x}\right)^3 \]
This simplifies to:
\[ T_4 = \binom{n}{3} p^{n-3} x^{n-3} \cdot x^{-3} = \binom{n}{3} p^{n-3} x^{n-6} \]

### Step 3: Condition for Independence of \(x\)
For the term \(T_4\) to be independent of \(x\), the exponent of \(x\) must be zero:
\[ n - 6 = 0 \implies n = 6 \]

### Step 4: Substitute \(n\) into the Fourth Term
Now substitute \(n = 6\) into the expression for \(T_4\):
\[ T_4 = \binom{6}{3} p^{6-3} = \binom{6}{3} p^3 \]

### Step 5: Calculate \(\binom{6}{3}\)
Calculating \(\binom{6}{3}\):
\[ \binom{6}{3} = \frac{6!}{3! \cdot 3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
Thus, we have:
\[ T_4 = 20 p^3 \]

### Step 6: Set the Fourth Term Equal to \(\frac{5}{2}\)
We know that the fourth term is equal to \(\frac{5}{2}\):
\[ 20 p^3 = \frac{5}{2} \]

### Step 7: Solve for \(p^3\)
To find \(p^3\), we rearrange the equation:
\[ p^3 = \frac{5}{2 \times 20} = \frac{5}{40} = \frac{1}{8} \]

### Step 8: Find \(p\)
Taking the cube root:
\[ p = \sqrt[3]{\frac{1}{8}} = \frac{1}{2} \]

### Final Answers
Thus, the value of \(n\) is \(6\) and the value of \(p\) is \(\frac{1}{2}\).