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JEE Main Math Class 11 Sequences and Series Part 1 Questions
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© examsnet.com
Question : 19
Total: 100
Let
1
x
1
,
1
x
2
,
1
x
3
,
.
.
.
.
,
(
x
i
≠
0
for
i
=
1
,
2
,
.
.
.
,
n
)
be in A.P.
such that
x
1
=
4
and
x
21
=
20
.
If
n
is the least positive integer for which
x
n
>
50
,
then
n
∑
i
=
1
(
1
x
i
)
is equal to.
[Online April 16, 2018]
3
13
8
13
4
1
8
Validate
Solution:
∵
1
x
1
,
1
x
2
,
1
x
3
,
.
.
.
.
,
1
x
n
are in A.P
x
1
=
4
and
x
21
=
20
Let
′
d
' be the common difference of this A.P
its
21
st
term
=
1
x
21
=
1
x
1
+
[
(
21
−
1
)
×
d
]
⇒
d
=
1
20
×
(
1
20
−
1
4
)
⇒
d
=
−
1
100
Also
x
n
>
50
(given).
∴
1
x
n
=
1
x
1
+
[
(
n
−
1
)
×
d
]
⇒
x
n
=
x
1
1
+
(
n
−
1
)
×
d
×
x
1
∴
x
1
1
+
(
n
−
1
)
×
d
×
x
1
>
50
⇒
4
1
+
(
n
−
1
)
×
(
−
1
100
)
×
4
>
50
⇒
1
+
(
n
−
1
)
×
(
−
1
100
)
×
4
<
4
50
⇒
−
1
100
(
n
−
1
)
<
−
23
100
⇒
n
−
1
>
23
⇒
n
>
24
Therefore,
n
=
25
.
⇒
25
∑
i
=
1
1
x
i
=
25
2
[
(
2
×
1
4
)
+
(
25
−
1
)
×
(
−
1
100
)
]
=
13
4
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