Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
If the constant term in the expansion of (3 x3-2 x2+(5/x5))10 is 2x . l, where l is an odd integer, then the value of k is equal to
Q. If the constant term in the expansion of
(
3
x
3
−
2
x
2
+
x
5
5
)
10
is
2
x
.
l
, where
l
is an odd integer, then the value of
k
is equal to
173
161
Report Error
A
6
B
7
C
8
D
9
Solution:
General term
T
r
+
1
=
∣
r
1
∣
r
2
∣
r
3
⌊
10
3
r
1
+
2
r
2
−
5
r
3
=
0
…………
.
(
1
)
r
1
+
r
2
+
r
3
=
10
…………
.
(
2
)
From equation (1) and (2)
r
1
+
2
(
10
−
r
3
)
−
5
r
3
=
0
r
1
+
20
=
7
r
3
(
r
1
,
r
2
,
r
3
)
=
(
1
,
6
,
3
)
Constant term
=
⌊
1∣6∣3
⌊
10
(
3
)
1
(
−
2
)
6
(
5
)
3
=
2
9
⋅
3
2
⋅
5
4
⋅
7
1
l
=
9