If ${(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is expanded in descending powers of x, then $7^{t h}$ term is given by:

7/18

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

16x/27

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

16/27

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

-(7x/18)

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

Open in App

Solution

The correct option is A
7/18

$T_{7} = T_{6 + 1} =^{9} C_{6} (\frac{3}{2} x^{2})^{3} (- \frac{1}{3 x})^{6}$ = $\frac{7}{18}$

Suggest Corrections

Similar questions

7 t h

{(\frac{1}{y} + y^{2})}^{10}

y

\frac{1}{x^{4} - 5 x^{3} + 7 x^{2} + x - 8}

x

The term independent of x in the expansion of $(1 + x + 2 x^{4}) (\frac{3}{2} x^{2} - \frac{1}{3 x})^{9}$ is

x

{(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}

x

{(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}

k

18 k

Join BYJU'S Learning Program