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JEE Main Math Class 11 Sequences and Series Part 3 Questions
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© examsnet.com
Question : 5
Total: 101
If
1
√
1
+
√
2
+
1
√
2
+
√
3
+
.
.
.
+
1
√
99
+
√
100
=
m
and
1
1
⋅
2
+
1
2
⋅
3
+
.
.
.
+
1
99
⋅
100
=
n
, then the point
(
m
,
n
)
lies on the line
[5 Apr 2024 Shift 1]
11
(
x
−
1
)
−
100
(
y
−
2
)
=
0
11
(
x
−
2
)
−
100
(
y
−
1
)
=
0
11
(
x
−
1
)
−
100
y
=
0
11
x
−
100
y
=
0
Validate
Solution:
👈: Video Solution
1
√
1
+
√
2
+
1
√
2
+
√
3
+
.
.
.
+
1
√
99
+
√
100
=
m
√
1
−
√
2
−
1
+
√
2
−
√
3
−
1
.
.
.
√
99
−
√
100
−
1
=
m
√
100
−
1
=
m
⇒
m
=
9
1
1
⋅
2
+
1
2
⋅
3
+
.
.
.
1
99
⋅
100
=
n
1
1
−
1
2
+
1
2
−
1
3
.
.
.
1
99
−
1
100
=
n
1
−
1
100
=
n
99
100
=
n
(
m
,
n
)
=
(
9
,
99
100
)
⇒
11
(
9
)
−
100
(
99
100
)
=
99
−
99
=
0
Ans. option (4)
11
x
−
100
∣
y
=
0
© examsnet.com
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