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JEE Main 6 Apr 2024 Shift 1 Solved Paper
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© examsnet.com
Question : 8
Total: 89
For
α
,
β
∈
R
and a natural number
n
, let
A
r
=
|
r
1
n
2
2
+
α
2
r
2
n
2
−
β
3
r
−
2
3
n
(
3
n
−
1
)
2
|
. Then
2
A
10
−
A
8
is
[6 Apr 2024 Shift 1]
4
α
+
2
β
2
α
+
4
β
2
n
0
Validate
Solution:
👈: Video Solution
A
r
=
|
r
1
n
2
2
+
α
2
r
2
n
2
−
β
3
r
−
2
3
n
(
3
n
−
1
)
2
|
2
A
10
−
A
8
=
|
20
1
n
2
2
+
α
40
2
n
2
−
β
56
3
n
(
3
n
−
1
)
2
|
-
−
|
8
1
n
2
2
+
α
16
2
n
2
−
β
22
3
n
(
3
n
−
1
)
2
|
⇒
|
12
1
n
2
2
+
α
24
2
n
2
−
β
34
3
n
(
3
n
−
1
)
2
|
⇒
|
0
1
n
2
2
+
α
0
2
n
2
−
β
−
2
3
n
(
3
n
−
1
)
2
|
⇒
−
2
(
(
n
2
−
β
)
−
(
n
2
+
2
α
)
)
⇒
−
2
(
−
β
−
2
α
)
⇒
4
α
+
2
β
© examsnet.com
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