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(A) For a binomial expansion with a rational index $n$,the general term is given by $T_{k+1} = \frac{n(n-1)(n-2)\dots(n-k+1)}{k!} x^k$. For the expans

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$101$

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(A) For a binomial expansion with a rational index $n$,the general term is given by $T_{k+1} = \frac{n(n-1)(n-2)\dots(n-k+1)}{k!} x^k$. For the expansion $(1+\frac{3x}{5})^{22/3}$,the $p^{	ext{th}}$ term is $T_p = \frac{n(n-1)\dots(n-p+2)}{(p-1)!} (\frac{3x}{5})^{p-1}$. The term becomes negative when the product $n(n-1)\dots(n-p+2) Since $n = 22/3 \approx 7.33$,the terms are positive as long as $n-k+1 > 0$. For $p^{	ext{th}}$ term,we need $n-(p-1)+1 $22/3 - p + 2  9.33$. Thus,$p=10$. For the expansion $(1-\frac{3x}{5})^{22/3}$,the general term is $T_k = \binom{n}{k-1} (-1)^{k-1} (\frac{3x}{5})^{k-1}$. For terms to be positive from $r^{	ext{th}}$ term onwards,the coefficient must be positive. Since $(-1)^{k-1}$ alternates,the condition requires the binomial coefficient part to be zero or the expansion to terminate. However,for non-integer $n$,the expansion is infinite. Re-evaluating the condition for $(1-\frac{3x}{5})^{22/3}$,the terms are $T_k = \frac{n(n-1)\dots(n-k+2)}{(k-1)!} (-1)^{k-1} (\frac{3x}{5})^{k-1}$. For $k \ge r$,the terms are positive if the sign of the coefficient matches $(-1)^{k-1}$. Following the same logic as $p$,we find $r=10$. Thus,the expansion is $(10x + \frac{10}{x})^{100}$. The number of terms in $(a+b)^n$ is $n+1$. Therefore,the number of terms is $100+1 = 101$.

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

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