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Byju's Answer
Standard XII
Mathematics
Continuity in an Interval
Let n be an i...
Question
Let n be an integer greater than 1 and let
a
n
=
1
log
n
1001
if
b
=
a
3
+
a
4
+
a
5
+
a
6
and
c
=
a
11
+
a
12
+
a
13
+
a
14
+
a
15
. Then value of
(
b
−
c
)
is equal to
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Solution
a
n
=
1
log
n
1001
log
x
P
=
log
y
P
log
y
x
b
=
a
3
+
a
4
+
a
5
+
a
6
=
1
log
3
1001
+
1
log
4
1001
+
1
log
5
1001
+
1
log
6
1001
=
log
2
3
log
2
1001
+
log
2
4
log
2
1001
+
log
2
5
log
2
1001
+
log
2
6
log
2
1001
=
log
2
(
3
×
4
×
5
×
6
)
log
2
1001
(
∵
log
m
+
log
n
=
log
m
n
)
=
log
2
(
360
)
log
2
1001
Similarly
c
=
a
11
+
a
12
+
a
13
+
a
14
+
a
15
=
log
2
(
11
×
12
×
13
×
14
×
15
)
log
2
(
1001
)
=
log
2
(
336336
)
log
2
(
1001
)
b
−
c
=
log
2
360
log
2
1001
−
log
2
(
336336
)
log
2
(
1001
)
log
2
(
360
×
336336
)
/
log
2
1001
(
∵
log
n
−
1
n
=
log
m
/
n
)
=
log
2
(
360
336336
)
log
2
(
1001
)
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0
Similar questions
Q.
Let
A
1
=
[
a
1
]
A
2
=
[
a
2
a
3
a
4
a
5
]
A
3
=
⎡
⎢
⎣
a
6
a
7
a
8
a
9
a
10
a
11
a
12
a
13
a
14
⎤
⎥
⎦
⋯
⋯
A
n
=
[
⋯
]
where
a
r
=
[
log
2
r
]
(
[
⋅
]
denotes the greatest integer
)
. Then trace of
A
10
is
Q.
Let
a
n
=
∫
π
4
0
tan
n
x
d
x
.
Then
a
2
+
a
4
,
a
3
+
a
5
,
a
4
+
a
6
are in
Q.
Let
a
n
=
∫
π
/
4
0
t
a
n
n
x
d
x
. Then
a
2
+
a
4
,
a
3
+
a
5
,
a
4
+
a
6
are in
Q.
Let
a
1
,
a
2
,
a
3
,
⋯
,
a
15
be in an A.P
a
1
+
a
2
+
⋯
+
a
10
=
A
a
6
+
a
7
+
⋯
+
a
15
=
B
If
B
−
A
=
200
and
B
+
A
=
860
,
then find
a
15
Q.
If
⎡
⎢
⎣
cos
θ
−
sin
θ
0
sin
θ
cos
θ
0
0
0
1
⎤
⎥
⎦
then the value of
a
11
A
11
+
a
12
A
12
+
a
13
A
13
=
where
A
11
,
A
12
A
13
are cofactors of
a
11
,
a
12
,
a
13
respectively
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