If the r th term in the expansion of x -3-2- x 210 contains X 4- there r is equal toA- 4B- 3C- 2D- None of these

The correct option is B 3T_r=T_(r 1)+1=^10C_r 1(x/3 )^10 (r 1)( 2/x^2 )^r 1 =^10C_r 1(1/3 )^11 4 ( 2)^r 1x^13 3r Let T_r+1contains x^4. ∴ 13 3r=4 i.e., r=3 ...

337

You visited us 337 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Progression

If the r th t...

Question

If the $r^{t h}$ term in the expansion of ${(\frac{x}{3} - \frac{2}{x^{2}})}^{10}$ contains $x^{4}$ , there r is equal to

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

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

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

None of these

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

Open in App

Solution

The correct option is B 3
$T_{r} = T_{(r - 1) + 1} =^{10} C_{r - 1} {(\frac{x}{3})}^{10 - (r - 1)} {(- \frac{2}{x^{2}})}^{r - 1} =^{10} C_{r - 1} {(\frac{1}{3})}^{11 - 4} (- 2)^{r - 1} x^{13 - 3 r} L e t T_{r + 1} c o n t a i n s x^{4} . ∴ 13 - 3 r = 4 i . e ., r = 3$

Suggest Corrections

Similar questions

Q. Which term in the expansion of

{(\frac{x}{3} - \frac{2}{x^{2}})}^{10}

contains

x^{4}

If the absolute term (independent of x) in the expansion of ${(\sqrt{x} - \frac{k}{x^{2}})}^{10}$ is 405 then k =

If rth term in the expansion of ${(2 x^{2} - \frac{1}{x})}^{12}$ is without x, then r is equal to

Q. If x⁴ occurs in the r^th terms in the expansion of

{(x^{4} + \frac{1}{x^{3}})}^{15},

then r = ___________.

Q. If r^th term is the middle term in the expansion of

{(x^{2} - \frac{1}{2 x})}^{20}

, then

{(r + 3)}^{t h}

term is
(a)

{}^{20}C_{14} (\frac{x}{2^{14}})

(b)

{}^{20}C_{12} x^{2} 2^{- 12}

(c)

- {}^{20}C_{7} x, 2^{- 13}

(d) none of these

Join BYJU'S Learning Program

If the rth term in the expansion of (x3−2x2)10 contains x4, there r is equal to

The correct option is B 3Tr=T(r−1)+1=10Cr−1(x3)10−(r−1)(−2x2)r−1=10Cr−1(13)11−4(−2)r−1x13−3rLet Tr+1contains x4.∴ 13−3r=4 i.e., r=3

If the $r^{t h}$ term in the expansion of ${(\frac{x}{3} - \frac{2}{x^{2}})}^{10}$ contains $x^{4}$ , there r is equal to

The correct option is B 3
$T_{r} = T_{(r - 1) + 1} =^{10} C_{r - 1} {(\frac{x}{3})}^{10 - (r - 1)} {(- \frac{2}{x^{2}})}^{r - 1} =^{10} C_{r - 1} {(\frac{1}{3})}^{11 - 4} (- 2)^{r - 1} x^{13 - 3 r} L e t T_{r + 1} c o n t a i n s x^{4} . ∴ 13 - 3 r = 4 i . e ., r = 3$