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Byju's Answer
Standard XII
Mathematics
Continuity of a Function
The G.M. of n...
Question
The G.M. of n positive terms
x
1
,
x
2
,
.
.
.
.
.
x
n
is
A
(
x
1
×
x
2
×
.
.
.
.
.
×
x
n
)
n
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B
1
n
(
x
1
×
x
2
×
.
.
.
×
x
n
)
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C
(
x
1
×
x
2
×
.
.
.
×
x
n
)
1
/
n
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D
None of these
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Solution
The correct option is
C
(
x
1
×
x
2
×
.
.
.
×
x
n
)
1
/
n
G.M of
n
positive number
x
1
,
x
2
,
x
3
,
.
.
.
.
.
.
.
x
n
is given by,
(
x
1
×
x
2
×
x
3
×
.
.
.
.
.
.
.
.
.
.
x
n
)
1
/
n
Suggest Corrections
0
Similar questions
Q.
If
x
1
,
x
2
,
x
3
,
x
4
.
.
.
.
x
2
n
+
1
are in Arithmetic Progression, then find the value of
[
(
x
2
n
+
1
−
x
1
)
(
x
2
n
+
1
+
x
1
)
]
+
[
(
x
2
n
−
x
2
)
(
x
2
n
+
x
2
)
]
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
[
(
x
n
+
2
−
x
n
)
(
x
n
+
2
+
x
n
)
]
Q.
Let
x
1
,
x
2
,
.
.
.
.
,
x
n
be the divisors of positive integer
′
n
′
(including
1
and
n
). If
x
1
+
x
2
+
.
.
.
.
+
x
n
=
75
, then
∑
i
=
1
1
x
i
is equal to
Q.
If
x
1
,
x
2
,
.
.
.
.
x
n
are
n
non-zero real numbers such that
(
x
2
1
+
x
2
2
+
x
2
3
+
.
.
.
.
.
x
2
n
+
1
)
(
x
2
2
+
x
2
3
+
x
2
4
+
.
.
.
.
x
2
n
)
≤
(
x
1
x
2
+
x
2
x
3
+
.
.
.
.
x
n
−
1
x
n
)
2
then prove that
x
1
,
x
2
,
.
x
n
are in
G
.
P
.
Q.
Standard deviation for n observations
x
1
,
x
2
,
…
x
n
is 5, then the standard deviation for n observations
x
1
+
1
,
x
2
+
1
,
…
x
n
+
1
will be
___
Q.
The sequence
(
x
n
)
n
≥
1
.
is definced by
x
1
=
1
,
x
2
=
3
and
x
n
+
2
=
6
x
n
+
1
−
x
n
,
for all
n
≥
1.
Then
x
2
+
(
−
1
)
n
is a perfect square, for all :
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