If $T_4$ represents the $4^{th}$ term in the expansion of $\left(5x + \frac{7}{x}\right)^{-3/2}$ and $x \notin \left[-\sqrt{\frac{7}{5}}, \sqrt{\frac{7}{5}}\right]$,then $\left(x^7 \sqrt{5x}\right) T_4 =$

$1 - \frac{1}{8} + \frac{1}{8} \cdot \frac{3}{16} - \frac{1 \cdot 3 \cdot 5}{8 \cdot 16 \cdot 24} + \dots =$

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

Assertion $(A) : 1+\frac{2}{3} \cdot \frac{1}{2}+\frac{2 \cdot 5}{3 \cdot 6} \cdot \frac{1}{4}+\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9} \cdot \frac{1}{8}+\ldots \infty = \sqrt[3]{4}$
Reason $(R) : |x| < 1, (1-x)^{-n} = 1+nx+\frac{n(n+1)}{1 \cdot 2} x^2+\frac{n(n+1)(n+2)}{1 \cdot 2 \cdot 3} x^3+\ldots$ The correct answer is

If $(a + bx)^{-2} = \frac{1}{4} - 3x + \dots$,then $(a, b) = $

$1+\frac{1}{4}+\frac{1 \cdot 3}{4 \cdot 8}+\frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12}+\ldots$ is equal to