The coefficient of $x^3$ in the expansion of $(1-\frac{3}{4} x)^{\frac{1}{2}}$ is

If $y = \frac{3}{4} + \frac{3 \times 5}{4 \times 8} + \frac{3 \times 5 \times 7}{4 \times 8 \times 12} + \ldots$ to $\infty$,then

If $x$ is numerically so small that $x^2$ and higher powers of $x$ can be neglected, then $\left(1+\frac{2x}{3}\right)^{3/2} \cdot (32+5x)^{-1/5}$ is approximately equal to

$1+\frac{2}{4}+\frac{2 \cdot 5}{4 \cdot 8}+\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\frac{2 \cdot 5 \cdot 8 \cdot 11}{4 \cdot 8 \cdot 12 \cdot 16}+\ldots \ldots$ is equal to :

If the coefficient of $x^{13}$ in the expansion of $\frac{(1+x)^2}{(1-2x)^3}$ is $A \times 2^{10}$,then $A=$

If the set of all values of $x$ for which the expansion of $(7-5 x)^{-\frac{2}{3}}$ is valid is equal to $(-a, a)$,then $5 a+7$ is equal to