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JEE Mains 11-Apr-2023 Shift 1 Solved Paper
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© examsnet.com
Question : 1
Total: 90
Let
x
1
,
x
2
,
.
.
.
.
.
,
x
100
be in an arithmetic progression, with
x
1
=
2
and their mean equal to 200 . If
y
i
=
i
(
x
i
−
i
)
,
1
≤
i
≤
100
, then the mean of
y
1
,
y
2
,
.
.
.
.
.
y
100
is :
[11-Apr-2023 shift 1]
10051.50
10100
10101.50
10049.50
Validate
Solution:
Mean
=
200
⇒
100
2
(
2
×
2
+
99
d
)
100
=
200
⇒
4
+
99
d
=
400
⇒
d
=
4
y
i
=
i
(
x
i
−
1
)
=
i
(
2
+
(
i
−
1
)
4
−
i
)
=
3
i
2
−
2
i
Mean
=
Σ
y
i
100
=
1
100
100
∑
i
=
1
3
i
2
−
2
i
=
1
100
{
3
×
100
×
101
×
201
6
−
2
×
100
×
101
2
}
=
101
{
201
2
−
1
}
=
101
×
99.5
=
10049.50
© examsnet.com
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