Find the middle term in the expansion of $\left(\frac{x}{3}+9 y\right)^{10}$.

Let $s_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$s_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $s_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.
Statement $-1$: $s_3 = 55 \times 2^9$
Statement $-2$: $s_1 = 90 \times 2^8$ and $s_2 = 10 \times 2^8$

The coefficient of the term independent of $x$ in the expansion of ${\left( {\sqrt {\frac{x}{3}} + \frac{3}{{2{x^2}}}} \right)^{10}}$ is

Let $n$ be a positive integer. If the coefficients of $2^{\text{nd}}$,$3^{\text{rd}}$,and $4^{\text{th}}$ terms in the expansion of $(1+x)^n$ are in $A$.$P$.,then the value of $n$ is:

The sum of all possible values of $n \in N$ such that the coefficients of $x$,$x^2$,and $x^3$ in the expansion of $(1+x^2)^2(1+x)^n$ are in arithmetic progression is:

Find the term independent of $x$ $(x > 0, x \neq 1)$ in the expansion of $\left[\frac{(x+1)}{\left(x^{2/3}-x^{1/3}+1\right)}-\frac{(x-1)}{(x-\sqrt{x})}\right]^{10}$.