The coefficient of $x^{-4}$ in the expansion of $( \frac {4x}{5} + \frac {5}{2x} )^8$ is?

$\textbf{Step 1: Find value of r to obtain given coefficient of x}$

$\text{Given equation is}$

$\bigg( \dfrac {4x}{5} + \dfrac {5}{2x}\bigg)^8$

$\text{General term of expansion $ (p+q)^n$ is given by-}$

$T_n= T_{r +1 }=^nC_r p^{ (n-r)}.q^r$

$\text{Comparing given equation with standard form, p=4x/5 and q=5/2x and n=8}$

$\Rightarrow T_n=T_{r+1}$

$= ^8C_r .\bigg( \dfrac {4x}{5}\bigg)^{(8-r)} .\bigg( \dfrac {5}{2x}\bigg)^r$

$= ^8C_r .\bigg( \dfrac {4}{5}\bigg)^{(8-r)} \times \bigg( \dfrac {5}{2}\bigg)^r \times x^{8-r-r}$

$= ^8C_r x^{(8 -2r)}. \bigg(\dfrac { 2^{(16 -3r)}}{5 ^{(8-2r)} }\bigg)$

$\textbf{Step 2: Find value of coefficient of x$^{-4}$}$

$\text{To obtain coefficient of x}^{-4} :$

$\Rightarrow 8-2r=-4$

$\Rightarrow$ $2r=12$

$\therefore$ $r=6$

$\text{For r=6,7th term is given by-}$

$T_7=T_{6+1}$

$=^8C_6 x^{-4} . \dfrac { 2^{ (16 -3 \times 6)} }{5^{(8-2\times6)} }$

$= ^8C_6 x^{-4} . \dfrac {5^4}{2^2 }$

$= \dfrac { 8 \times 7}{ 2 \times 1} \times \dfrac {625}{4} x^{-4} .$

$= 4375 x^{-4}$

$\therefore$ $\text{Coefficient of } x^{-4}$$=4375$

$\textbf{Therefore,value of coefficient of x$^{-4}$ is 462}$