If the $6^{th}$ term in the expansion of the binomial ${\left[ {\sqrt {{2^{\log (10 - {3^x})}}} + \sqrt[5]{{{2^{(x - 2)\log 3}}}}} \right]^m}$ is equal to $21$ and it is known that the binomial coefficients of the $2^{nd}$, $3^{rd}$ and $4^{th}$ terms in the expansion represent respectively the first, third and fifth terms of an $A.P$. (the symbol log stands for logarithm to the base $10$), then $x = $

Find an approximation of $(0.99)^{5}$ using the first three terms of its expansion.

Find $a, b$ and $n$ in the expansion of $(a+b)^{n}$ if the first three terms of the expansion are $729,7290$ and $30375,$ respectively.

If $\alpha$ and $\beta$ be the coefficients of $x^{4}$ and $x^{2}$ respectively in the expansion of

$(\mathrm{x}+\sqrt{\mathrm{x}^{2}-1})^{6}+(\mathrm{x}-\sqrt{\mathrm{x}^{2}-1})^{6}$, then 

The sum of the coefficients of the first three terms in the expansion of $\left(x-\frac{3}{x^{2}}\right)^{m}, x \neq 0, m$ being a natural number, is $559 .$ Find the term of the expansion containing $x^{3}$

The coefficient of $x^4$ in ${\left[ {\frac{x}{2}\,\, - \,\,\frac{3}{{{x^2}}}} \right]^{10}}$ is :