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JEE Mains 10-Apr-2023 Shift 2 Solved Paper
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© examsnet.com
Question : 9
Total: 90
If
S
n
=
4
+
11
+
21
+
34
+
50
+
.
.
.
. to
n
terms, then
1
60
(
S
29
−
S
9
)
is equal to
[10-Apr-2023 shift 2]
220
227
226
223
Validate
Solution:
S
n
=
4
+
11
+
21
+
34
+
50
+
.
.
.
+
n
terms
Difference are in A.P.
Let
T
n
=
a
n
2
+
b
n
+
c
T
1
=
a
+
b
+
c
=
4
T
2
=
4
a
+
2
b
+
c
=
11
T
3
=
9
a
+
3
b
+
c
=
21
By solving these 3 equations
a
=
3
2
,
b
=
5
2
,
c
=
0
So
T
n
=
3
2
n
2
+
5
2
n
S
n
=
Σ
T
n
=
3
2
∑
n
2
+
5
2
∑
n
=
3
2
n
(
n
+
1
)
(
2
n
+
1
)
6
+
5
2
(
n
)
(
n
+
1
)
2
=
n
(
n
+
1
)
4
[
2
n
+
1
+
5
]
S
n
=
n
(
n
+
1
)
4
(
2
n
+
6
)
=
n
(
n
+
1
)
(
n
+
3
)
2
1
60
(
29
×
30
×
32
2
−
9
×
10
×
12
2
)
=
223
© examsnet.com
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