To solve the problem, let's first recall the general form of the binomial expansion:
[tex]\[
\left(a + b\right)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
In this case, the given binomial expression is [tex]\(\left(px + \frac{1}{x}\right)^n\)[/tex]. Here, [tex]\(a = px\)[/tex] and [tex]\(b = \frac{1}{x}\)[/tex]. According to the problem, the fourth term of this expansion equals [tex]\(\frac{5}{2}\)[/tex].
To find the fourth term in the binomial expansion, we use the formula for the binomial term [tex]\(T_{k+1}\)[/tex]:
[tex]\[
T_{k+1} = \binom{n}{k} a^{n-k} b^k
\][/tex]
For the fourth term ([tex]\(T_4\)[/tex]), we set [tex]\(k = 3\)[/tex]:
[tex]\[
T_4 = \binom{n}{3} (px)^{n-3} \left(\frac{1}{x}\right)^3
\][/tex]
Now, we simplify [tex]\(T_4\)[/tex]:
[tex]\[
T_4 = \binom{n}{3} (px)^{n-3} \cdot \frac{1}{x^3} = \binom{n}{3} p^{n-3} x^{n-3-3} x^{-3} = \binom{n}{3} p^{n-3} x^{n-6}
\][/tex]
Given that [tex]\(T_4\)[/tex] must equal [tex]\(\frac{5}{2}\)[/tex], we ignore [tex]\(x^{n-6}\)[/tex] since it cancels out, and we get:
[tex]\[
\binom{n}{3} p^{n-3} = \frac{5}{2}
\][/tex]
We evaluate this expression to identify [tex]\(n\)[/tex] and [tex]\(p\)[/tex], and this gives us the exact terms. Having explored the options [tex]\(n \in \{6, 8\}, p \in \{6, \frac{1}{2}\}\)[/tex], we find that the successful combination of [tex]\(n\)[/tex] and [tex]\(p\)[/tex] must satisfy the equation exactly.
Upon thorough evaluation and substitution, we find that:
For [tex]\(n = 6\)[/tex] and [tex]\(p = \frac{1}{2}\)[/tex],
[tex]\(\binom{6}{3}\left(\frac{1}{2}\right)^{6-3}\)[/tex]
We know:
[tex]\(\binom{6}{3} = 20\)[/tex] and [tex]\(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\)[/tex],
So,
[tex]\(20 \times \frac{1}{8} = \frac{20}{8} = \frac{5}{2}\)[/tex]
Therefore, the correct answer is:
[tex]\[
\boxed{d) \, n=6, \, p=\frac{1}{2}}
\][/tex]
