${\left( {\frac{x}{2} – \frac{3}{{{x^2}}}} \right)^{10}}$ के प्रसार में व्यापक पद =

${T_{r + 1}} = {\,^{10}}{C_r}{\left( {\frac{x}{2}} \right)^{10 – r}}.\,\,{\left( { – \frac{3}{{{x^2}}}} \right)^r}$

  ${ = ^{10}}{C_r}{( – 1)^r}.\frac{{{3^r}}}{{{2^{10 – r}}}}{x^{10 – r – 2r}}$

यहाँ $x$ की घात $4$ है। अत: $10 – 3r = 4 \Rightarrow r = 2$

$\therefore \,\,\,\,{T_{2 + 1}}{ = ^{10}}{C_2}{\left( {\frac{x}{{\rm{2}}}} \right)^8}{\left( { – \frac{3}{{{x^2}}}} \right)^2} = \frac{{10.9}}{{1.2}}.\frac{1}{{{2^8}}}{.3^2}.{x^4}$

= $\frac{{405}}{{256}}{x^4}$

$\therefore $ अभीष्ट गुणांक $ = \frac{{405}}{{256}}$.