The numerically greatest term in the expansion ${(\text{5}x–\text{ 6}y)}^{\text{14}}$ when $x=\frac{2}{5},y=\frac{1}{2}$ is

Given expansion ${(\text{5}x–\text{ 6}y)}^{\text{14}}$ is

${\left(5x-6y\right)}^{14}={\left(5x\right)}^{14}{\left(1-\frac{6y}{5x}\right)}^{14}$  this expansion covert into ${\left(1+x\right)}^n$

The numerically greatest term in the expansion of${\left(1+x\right)}^n$ is $\frac{\left(n+1\right)\left|x\right|}{\left|x\right|+1}$

$N.G.T=\frac{15\left|\frac{6\times \frac{1}{2}}{5\times \frac{2}{5}}\right|}{\frac{3}{2}+1}=9$

 We have general term in the expansion ${\left(x+a\right)}^n$

$(\therefore T_{r+1}={}^n{\text{C}}_r{x}^{n-r}{\left(a\right)}^r$ be the expansion of ${(x+a)}^n)$

$T_9=T_{8+1}{=}^{14}{\text{C}}_8{\left(5x\right)}^6{\left(-6y\right)}^8\mathrm{........................}\left(1\right)$

Substitute these values in Eqn (1)$x=\frac{2}{5},y=\frac{1}{2}$  then we get

      $T_9{=}^{14}{\text{C}}_8{2}^6\times 3^8$ is the numerically greatest term in the given expansion