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Test Index
JEE Main Math Class 11 Sequences and Series Part 3 Questions
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© examsnet.com
Question : 10
Total: 101
Let the first term of a series be
T
1
=
6
and its
r
th
term
T
r
=
3
T
r
−
1
+
6
r
,
r
=
2
,
3
,
.
.
.
.
.
.
,
n
. If the sum of the first
n
terms of this series is
1
5
(
n
2
−
12
n
+
39
)
(
4.6
n
−
5.3
n
+
1
)
. Then
n
is equal to ______.
[6 Apr 2024 Shift 1]
Your Answer:
Validate
Solution:
👈: Video Solution
T
r
=
3
T
r
−
1
+
6
r
,
r
=
2
,
3
,
4
,
.
.
.
n
T
2
=
3
⋅
T
1
+
6
2
T
2
=
3
⋅
6
+
6
2
.
.
.
(
1
)
T
3
=
3
T
2
+
6
3
T
3
=
3
T
2
+
6
3
T
3
=
3
(
3
⋅
6
+
6
2
)
+
6
3
T
3
=
3
2
⋅
6
+
3
⋅
6
2
+
6
3
.
.
.
(
2
)
T
r
=
3
r
−
1
⋅
6
+
3
r
−
2
⋅
6
2
+
.
.
.
+
6
r
T
r
=
3
r
−
1
⋅
6
[
1
+
6
3
+
(
6
3
)
2
+
.
.
.
+
(
6
3
)
r
−
1
]
T
r
=
3
r
−
1
⋅
6
(
1
+
2
+
2
2
+
.
.
.
+
2
r
−
1
)
T
r
=
6
⋅
3
r
−
1
1
⋅
(
1
−
2
r
)
(
−
1
)
T
r
=
6
⋅
3
r
−
1
⋅
(
2
r
−
1
)
T
r
=
6
⋅
3
r
3
⋅
(
2
r
−
1
)
T
r
=
2
⋅
(
6
r
−
3
r
)
S
n
=
2
Σ
(
6
r
−
3
r
)
S
n
=
2
⋅
[
6
⋅
(
6
n
−
1
)
5
−
3
⋅
(
3
n
−
1
)
2
]
S
n
=
2
[
12
(
6
n
−
1
)
−
15
(
3
n
−
1
)
10
]
S
n
=
3
5
[
4
⋅
6
4
−
5
⋅
3
n
+
1
]
∴
n
2
−
12
n
+
39
=
3
n
2
−
12
n
+
36
=
0
n
=
6
© examsnet.com
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