Formula Used:

General Term in the expansion (x - y)n = Tr+1 = (-1)r × nCr × xn-r ×  yr

Concept:

The middle term of the expansion of  (x + a)n depends upon the value of n.

(i) When n is even the total number of terms in the expansion of (x + a)n is n+1(odd).

So, there is only one middle term, i.e \(\rm\left ( \frac{n+1}{2} \right)\)th term is the middle term.

(ii) When n is odd The total number of terms in the expansion of (x + a)n is n+1(even).

So, there is two middle term, i.e., \(\rm\left ( \frac{n+1}{2} \right )\)th and \(\rm\left ( \frac{n+3}{2} \right )\)th term are two middle terms.

Calculation:

The number of terms in the expansion of \(\rm \left ( 4x-\frac{x^3}{2} \right ) ^9\) is 10(even).

So there are two middle terms. i.e. \(\rm\left ( \frac{9+1}{2} \right )\)th and \(\rm\left ( \frac{9+3}{2} \right )\) th  terms.

They are given by T₅ and T₆.

Tr+1 = nCrxn-rar

Required term 

T5 = T4+1 = 9C₄(4x)⁵(-x³/2)⁴

\(\Rightarrow \rm T_{5}= {}^9C_{4}4^5x^5\left ( \frac{1}{2^4} \right )x^{12}\)

\(\Rightarrow \rm T_{5}= {}^9C_{4}4^3x^{17}\)

T6 = T5+1 = 9C₅(4x)4(-x3/2)5

\(\Rightarrow \rm T_{6}= -{}^9C_{5}4^4x^4\left ( \frac{1}{2^5} \right )x^{15}\)

\(\Rightarrow \rm T_{6}=- {}^9C_{5}\left ( \frac{2^8}{2^5} \right )x^{19}\)

\(\Rightarrow \rm T_{6}=- {}^9C_{5}\ 2^3\ x^{19}\)

∴ The 3rd option is correct option.