If the constant term in the expansion of ( 3 x 2 - 1 2 x 5 ) 7 is, then is equal to _______ . [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
If the constant term in the expansion of ${(3 x^{2} - \frac{1}{2 x^{5}})}^{7}$ is $α$ , then $[α]$ is equal to _______ . [2023]

Ans.
(1275)

Let the $r^{t h}$ term be the constant term of the given expansion.

$∴$ $T_{r + 1} = {}^{7}{C_{r}} {(3 x^{2})}^{7 - r} \cdot {(\frac{- 1}{2 x^{5}})}^{r} = {}^{7}{C_{r}} 3^{7 - r} \cdot {(\frac{- 1}{2})}^{r} \cdot x^{14 - 2 r - 5 r}$

For constant term, put $14 - 2 r - 5 r = 0 \Rightarrow 7 r = 14 \Rightarrow r = 2$

$∴$ $T_{2 + 1} = {}^{7}{C_{2}} 3^{7 - 2} {(\frac{- 1}{2})}^{2} = 21 \times 3^{5} \times \frac{1}{4}$

$\Rightarrow α = 1275.75 \Rightarrow [α] = 1275$

Want Us to Email you About Special Offers & Updates?