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JEE Main Math Class 11 Sequences and Series Part 1 Questions
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© examsnet.com
Question : 6
Total: 100
Let
S
n
=
1
+
q
+
q
2
+
.
.
.
.
+
q
n
and
T
n
=
1
+
(
q
+
1
2
)
+
(
q
+
1
2
)
2
+
.
.
.
+
(
q
+
1
2
)
n
where
q
is a real number and
q
≠
1
. If
101
C
1
+
101
C
2
⋅
S
1
+
.
.
.
.
+
101
C
101
⋅
S
100
=
α
T
100
,
then
α
is equal to :
[Jan. 11, 2019 (II)]
2
99
202
200
2
100
Validate
Solution:
S
n
=
(
1
−
q
n
+
1
1
−
q
)
,
T
n
=
1
−
(
q
+
1
2
)
n
+
1
1
−
(
q
+
1
2
)
⇒
T
100
=
1
−
(
q
+
1
2
)
101
1
−
(
q
+
1
2
)
S
n
=
1
1
−
q
−
q
n
+
1
1
−
q
,
T
100
=
2
101
−
(
q
+
1
)
101
2
100
(
1
−
q
)
Now,
101
C
1
+
101
C
2
S
1
+
101
C
3
S
2
+
.
.
.
+
101
C
101
S
100
=
(
1
1
−
q
)
(
101
C
2
+
.
.
.
+
101
C
101
)
−
1
1
−
q
(
101
C
2
q
2
+
101
C
3
q
3
+
.
.
.
+
101
C
101
q
101
)
+
101
=
1
1
−
q
(
2
101
−
1
−
101
)
−
(
1
1
−
q
)
(
(
1
+
q
)
101
−
1
−
101
C
1
q
)
+
101
=
1
1
−
q
[
2
101
−
102
−
(
1
+
q
)
101
+
1
+
101
q
]
+
101
=
1
1
−
q
[
2
101
−
101
+
101
q
−
(
1
+
q
)
101
]
+
101
=
(
1
1
−
q
)
[
2
101
−
(
1
+
q
)
101
]
=
2
100
T
100
Hence, by comparison
α
=
2
100
© examsnet.com
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