The numerically greatest terms in the expansion of ${ \left( 2x+5y \right) ; }^{ 34 }$& ;when $x=3$ & $y=2$ is

$(2x+5y)^{34}$ can be written as $(2x)^{34}$.$(1+ \dfrac{5y}{2x})^{34}$

$\rightarrow$ or

$(2x)^{34}$.$(1+ \alpha)^{34}$, where $\alpha$ = $\dfrac{5y}{3x}$

Value of $\alpha$ at x=3, y=2 is equal to $\dfrac{5.2}{2.3}=\dfrac{5}{3}$

So $\alpha_{x=3,y=2}= \dfrac{5}{3}$

Now the numerically greatest term in the expansion of

$(2x+5y)^{34}$

is the numerically greatest term in the expansion of $(1+\alpha)^{34}$

We know that for any expansion $(1+\alpha)^n$, If the numerically greatest term is $T_r$, where r is an integer.

Then the value of r is $\leq\dfrac {(n+1)|\alpha|}{|\alpha+1|}$

So to know the numerically greatest term in the given expansion at x=3, y=2 we need to find the

integer value of

r

$\leq\dfrac {(n+1)|\alpha|}{|\alpha+1|}$ at x=3, y=2, Here

the value of n = 34

So $r\leq\dfrac{(34+1)|\dfrac{5}{3}|}{|\dfrac{5}{3}+1|}$

$\rightarrow$ $\dfrac{35.5}{8} = \dfrac{175}{8} = 21.8$

As r is an integer so the closest value of integer less than 21.8 is 21.

Hence the numerically greatest term will be 21th term or $T_{21}$

So the correct option is A.