When x = 1, $y = \frac{3}{2}$ , numerically greatest term in the expansion of $(2 x - 3 y)^{12}$ is

6 t h

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8 t h

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10 t h

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

12 t h

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $10 t h$
The numerically greatest term in $(1 + x)^{n}$ will be given by.
$\frac{(1 + n) (| x |)}{1 + | x |}$

$(2 x - 3 y)^{12}$
For $x = 1$ and $y = \frac{3}{2}$ ,
$(2 - \frac{9}{2})^{12}$
$= 2^{12} (1 - \frac{9}{4})^{12}$
Therefore numerically greatest term will be given by.
$\frac{13 (\frac{9}{4})}{1 + \frac{9}{4}}$
$= \frac{13 (9)}{13}$
$= 9$
$r = 9$ or the term is $10^{t h}$ term.

Suggest Corrections

Similar questions

The numerically greatest term in the expansion of $(2 x - 3 y)^{12}$ ) when $x = 1, y = \frac{5}{3}$ , is

{(2 x - 3 y)}^{12}

x = 1

y = 5 / 2

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

x = 3

y = 2

x = 9, y = 4

(2 x - 3 y)^{28}

(2 x + y)^{14}

x = 4

y = 5

Join BYJU'S Learning Program