For $x>0,$ if $p^{\text{th }}$term is the first negative term in the expansion of ${\left(1+\frac{3x}{5}\right)}^{22/3}$and in the expansion of ${\left(1-\frac{3x}{5}\right)}^{22/3}$ from $r^{th}$term onwards all the terms are positive, then the number of terms in the expansion of ${\left(px+\frac{r}{x}\right)}^{pr}$ is

For negative/rational index $n$,

${\left(1+x\right)}^n=1+nx+\frac{n\left(n-1\right)}{2!}+.....+\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)..................\left(n-r+1\right)}{r!}{x}^r+....$

In the given expansion ${\left(1+\frac{3x}{5}\right)}^{22/3}$, $p^{th}$ term is first negative term.

$\therefore T_p=\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)..................\left(n-p+2\right)}{\left(p-1\right)!}{x}^{p-1}\left[\text{Here}, n=\frac{22}{3}\right]$

Here, $n-p+2<0$

$\Rightarrow \frac{22}{3}-p+2<0$

$\Rightarrow -p<-\frac{28}{3}$

$\Rightarrow p>9.33$

$\therefore p=10$

Also in the expansion of ${\left(1-\frac{3x}{5}\right)}^{22/3}$ from $r^{th}$ term onwards all terms are positive.

$\therefore T_r={\left(-1\right)}^{r-1}\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)..................\left(n-r+2\right)}{\left(r-1\right)!}{x}^{r-1}\left[\text{Here}, n=\frac{22}{3}\right]$

$\therefore n-r+2<0 \& r-1$ is odd.

$\Rightarrow \frac{22}{3}-r+2<0$

$\Rightarrow r>9.33$

$\therefore r=10$ 

Now, In the expansion ${\left(px+\frac{r}{x}\right)}^{pr} \text{i.e.}$ ${\left(10x+\frac{10}{x}\right)}^{100}$,

So, number of terms$=100+1=101$.