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Byju's Answer
Standard XII
Mathematics
Logarithms
1+3+5+...+n4+...
Question
1
+
3
+
5
+
.
.
.
+
n
4
+
7
+
10
+
.
.
.
.
.
+
n
=
20
7
log
10
x
and n =
log
10
x
+
log
10
x
1
/
2
+
log
10
x
1
/
4
+
.
.
.
.
, then x is equal to
A
10
3
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B
10
5
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C
10
7
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D
10
9
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Solution
The correct option is
B
10
5
n
=
log
x
+
1
2
log
x
+
1
4
log
x
+
.
.
.
.
.
.
.
.
.
n
=
(
log
x
)
(
1
+
1
2
+
.
.
.
.
.
)
n
=
(
log
x
)
(
1
1
−
1
2
)
=
2
l
o
g
10
x
Now, in numerator,
1
+
3
+
.
.
.
.
.
(
n
t
e
r
m
s
)
=
∑
n
k
=
1
(
2
k
−
1
)
=
2
(
n
(
n
+
1
)
2
)
−
n
4
+
7
+
10
+
.
.
.
.
.
.
(
n
t
e
r
m
s
)
=
∑
n
k
=
1
(
3
k
+
1
)
=
3
n
(
n
+
1
)
2
+
n
Now,
n
2
n
2
(
3
n
+
5
)
=
20
7
log
10
x
2
n
3
n
+
5
=
20
7
log
10
x
l
e
t
log
10
x
=
z
∴
n
=
2
z
28
z
2
=
20
(
6
z
+
5
)
7
z
2
−
30
z
−
25
=
0
∴
z
=
5
log
10
x
=
5
x
=
10
5
Suggest Corrections
0
Similar questions
Q.
If
1
+
3
+
5
+
…
upto
n
terms
4
+
7
+
10
+
…
upto
n
terms
=
20
67
log
10
x
20
7
log
10
x
and
n
=
log
10
x
+
log
10
x
1
/
2
+
log
10
x
1
/
4
+
…
.
.
∞
,
then
x
is equal to
Q.
If for
x
,
y
∈
R
,
x
>
0
,
y
=
log
10
x
+
log
10
x
1
/
3
+
log
10
x
1
/
9
+
…
upto
∞
terms and
2
+
4
+
6
+
…
+
2
y
3
+
6
+
9
+
…
+
3
y
=
4
log
10
x
, then the ordered pair
(
x
,
y
)
is equal to:
Q.
Solve the following equations for
x
and
y
.
log
10
x
+
1
2
log
10
x
+
1
4
log
10
x
+
.
.
.
=
y
and
1
+
3
+
5
+
.
.
.
.
+
(
2
y
−
1
)
4
+
7
+
10
+
.
.
.
.
+
(
3
y
+
1
)
=
20
7
log
10
x
Q.
For a fixed
+
i
v
e
integer
n
, let
D
=
∣
∣ ∣ ∣
∣
(
n
−
1
)
!
(
n
+
2
)
!
(
n
+
3
)
!
/
n
(
n
+
1
)
!
(
n
+
1
)
!
(
n
+
3
)
!
(
n
+
5
)
!
/
(
n
+
2
)
!
(
n
+
3
)
!
(
n
+
3
)
!
(
n
+
5
)
!
(
n
+
7
)
!
/
(
n
+
4
)
!
(
n
+
5
)
!
∣
∣ ∣ ∣
∣
then
D
(
n
−
1
)
!
(
n
+
1
)
!
(
n
+
3
)
!
is equal to
Q.
Let
n
1
<
n
2
<
n
3
<
n
4
<
n
5
be positive integers such that
n
1
+
n
2
+
n
3
+
n
4
+
n
5
=
20.
The number of such distinct arrangements
(
n
1
,
n
2
,
n
3
,
n
4
,
n
5
)
is ?
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