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CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 9 - MCQExams.com
CBSE
Class 11 Engineering Maths
Sequences And Series
Quiz 9
The value of
∞
∑
n
=
2
(
1
−
1
n
2
)
equals
Report Question
0%
−
l
n
3
0%
0
0%
−
l
n
2
0%
−
l
n
5
solve that :-
Report Question
0%
14
0%
18
0%
11
0%
26
Explanation
Given,
Image
A
4
+
2
+
5
+
3
=
14
×
2
=
28
Image
B
7
+
5
+
4
+
3
=
19
×
2
=
38
∴
Image
C
2
+
1
+
3
+
7
=
13
×
2
=
26
The numbers
log
180
12
,
log
2160
12
,
log
25920
12
are in
Report Question
0%
AP
0%
GP
0%
HP
0%
None of these
If
[
x
]
denotes the greates integer
≤
x
, then
[
2
3
]
+
[
2
3
+
1
99
]
+
[
2
3
+
2
99
]
+
.
.
.
.
+
[
2
3
+
98
99
]
=
Report Question
0%
99
0%
98
0%
66
0%
65
Explanation
[
2
3
]
=
0
[
2
3
+
1
99
]
=
0
[
2
3
+
2
99
]
=
0
.
.
.
.
.
.
[
2
3
+
33
99
]
=
[
2
3
+
1
3
]
=
[
1
]
=
1
[
2
3
+
34
99
]
=
[
1
+
1
99
]
=
1
[
2
3
+
98
99
]
=
[
1
+
65
99
]
=
1
So, the sum is
1
+
1
+
1
+
.
.
.
.
.
.
.
.
.
66
times
=
66
If
(
1
+
a
x
)
n
=
1
+
8
x
+
24
x
2
+
.
.
.
;
then
a
−
n
a
+
n
is equal to
Report Question
0%
3
0%
−
1
3
0%
−
3
0%
1
3
Explanation
(
1
+
a
x
)
n
=
1
+
n
(
a
x
)
+
n
C
2
(
a
x
)
2
+
n
C
3
(
a
x
)
3
.
.
.
.
According to question
(
1
+
a
x
)
n
=
1
+
8
x
+
24
x
2
+
.
.
.
⇒
a
n
=
8
a
2
=
(
n
)
(
n
−
1
)
2
=
24
⇒
a
2
(
n
)
(
n
−
1
)
=
48
⇒
a
(
n
−
1
)
=
6
⇒
n
n
−
1
=
8
6
=
4
3
⇒
n
=
4
,
⇒
a
=
2
∴
a
−
n
a
+
n
=
2
−
4
2
+
4
=
−
2
6
=
−
1
3
If
S
=
1
2
−
2
2
+
3
2
−
4
2
.
.
.
.
upto
n
terms and
n
is even, then
S
equals _____
Report Question
0%
n
(
n
+
1
)
2
0%
n
(
n
−
1
)
2
0%
−
n
(
n
+
1
)
2
0%
−
n
(
n
−
1
)
2
Explanation
S
=
1
2
−
2
2
+
3
2
−
4
2
.
.
.
.
.
.
.
.
.
upto
n
term
=
(
1
−
2
)
(
1
+
2
)
+
(
3
−
4
)
(
3
+
4
)
+
.
.
.
.
.
.
+
(
n
−
1
−
n
)
(
n
−
1
+
n
)
=
−
(
1
+
2
+
3
+
4
+
.
.
.
.
.
+
n
)
S
=
−
n
(
n
+
1
)
2
The sum
m
∑
i
=
0
(
10
i
)
(
20
m
−
i
)
(where
(
p
q
)
=
0
if
p
<
q
) is maximum where
m
is
Report Question
0%
5
0%
10
0%
15
0%
20
The series
1
2
−
2
2
+
3
2
−
4
2
+
.
.
.
.
.
+
99
2
−
100
2
=
_____
Report Question
0%
−
5050
0%
5050
0%
11000
0%
−
11000
Explanation
The sum of nth term of the series is given as
S
=
(
−
1
)
n
+
1
n
(
n
+
1
)
2
Here
n
=
100
So,
S
=
(
−
1
)
100
+
1
100
(
100
+
1
)
2
=
(
−
1
)
10100
2
=
−
5050
If
x
and
y
are the number of possibilities that
A
can assume such that the unit digit of A and
A
3
are same and the unit digit of
A
2
and
A
3
are same respectively ,then the value of
x
−
y
is (where
A
is a single digit number)
Report Question
0%
4
0%
2
0%
3
0%
5
Explanation
Here, according to the question
if
A
=
1
, then
A
3
=
1
,
We see that unit digit number of
A
and
A
3
are same.
for,
A
=
2
⇒
A
3
=
8
,
, unit digits are not same
A
=
3
⇒
A
3
=
27
,
unit digits are not same
A
=
4
⇒
A
3
=
64
,
unit digits are same
A
=
5
⇒
A
3
=
125
,
unit digits are same
A
=
6
⇒
A
3
=
216
,
unit digits are not same
A
=
7
⇒
A
3
=
343
,
unit digits are not same
A
=
8
⇒
A
3
=
512
,
unit digits are not same
A
=
8
⇒
A
3
=
512
,
unit digits are same
So, there are five possible solutions where unit digits of
A
and
A
3
are same.
∴
x
=
5
Again, if we take
A
=
1
,
A
2
=
1
,
A
3
=
1
, unit digits are same
A
=
2
⇒
A
2
=
4
,
A
3
=
8
,
unit digits are not same
A
=
3
⇒
A
2
=
9
,
A
3
=
27
,
unit digits are not same
A
=
4
⇒
A
2
=
16
,
A
3
=
64
,
unit digits are not same
A
=
5
⇒
A
2
=
25
,
A
3
=
125
,
unit digits are same
A
=
6
⇒
A
2
=
36
,
A
3
=
216
,
unit digits are same
A
=
7
⇒
A
2
=
49
,
A
3
=
343
,
unit digits are not same
A
=
8
⇒
A
2
=
64
,
A
3
=
512
,
unit digits are not same
A
=
9
⇒
A
2
=
81
,
A
3
=
729
unit digits are not same
Here, we can see that there are three possible solutions where unit digits of
A
2
,
A
3
are same.
So,
y
=
3
Hence,
x
−
y
=
5
−
3
=
2.
[
12
47
21
10
52
4
64
?
24
]
Report Question
0%
40
0%
83
0%
62
0%
16
Explanation
[
12
47
21
10
52
4
64
?
24
]
We can see that dividing third number in each row with second digit of middle number and multiplying with first digit gives first member.
Similarly, out of
4
options, dividing
24
by
3
and multiplying by
8
gives
64
.
⇒
Option
(
B
)
.
In a triangle
A
B
C
,
a
c
o
s
2
(
C
2
)
+
c
c
o
s
2
(
A
2
)
=
3
b
2
, then the sides
a
,
b
,
c
Report Question
0%
Satisfy
a
+
b
=
c
0%
are in
A
.
P
.
0%
are in
G
.
P
.
0%
are in
H
.
P
.
Explanation
a
cos
2
(
C
2
)
+
C
cos
2
(
A
2
)
=
3
b
2
a
[
1
+
cos
C
2
]
+
C
[
1
+
cos
A
2
]
=
3
b
2
a
2
+
c
2
+
1
2
[
a
cos
C
+
C
cos
A
]
=
3
b
2
a
+
c
+
a
cos
C
+
c
cos
A
=
3
b
2
R
sin
A
+
2
R
sin
C
+
2
R
sin
A
cos
C
+
2
R
sin
C
cos
A
=
3
(
2
R
)
(
sin
B
)
sin
A
+
sin
C
+
sin
(
A
+
C
)
=
3
sin
(
B
)
A
+
B
+
C
=
π
⇒
sin
(
A
+
C
)
=
sin
B
sin
A
+
sin
C
+
sin
B
=
3
sin
(
B
)
sin
A
+
sin
C
=
2
sin
B
2
R
sin
A
+
2
R
sin
C
=
2
R
sin
B
a
+
c
=
2
b
∴
a
,
b
,
c
are an
A
.
P
The value of
n
−
1
∑
r
=
1
sin
2
r
π
n
is equal to
Report Question
0%
n
0%
n
2
0%
n
+
1
0%
Z
e
r
o
The sum of the series
10.
n
C
0
+
10
2
.
n
C
1
+
10
3
.
n
C
2
+
.
.
.10
n
+
1
.
n
C
n
is
Report Question
0%
11
n
0%
10.11
n
0%
11
n
+
1
0%
11
n
−
1
Explanation
→
10
(
n
C
0
+
10.
n
C
1
+
.
.
.
+
10
n
n
C
n
)
=
10
(
n
∑
r
=
1
n
C
r
(
10
)
r
(
1
)
n
−
r
)
=
10
(
1
+
10
)
n
=
10.11
n
=
(
B
)
1
+
(
1
+
a
)
x
+
1
(
1
+
a
+
a
2
)
x
2
+
(
1
+
a
+
a
2
+
a
3
)
x
3
+
−
−
−
−
−
where
0
<
a
,
x
<
1
,
is
−
Report Question
0%
1
(
1
−
x
)
(
1
−
a
)
0%
1
(
1
−
x
)
(
1
−
a
x
)
0%
1
(
1
−
a
)
(
1
−
a
x
)
0%
None of these
Explanation
We have,
1
+
(
1
+
a
)
x
+
1
(
1
+
a
+
a
2
)
x
2
+
(
1
+
a
+
a
2
+
a
3
)
x
3
+
.
.
.
.
.
.
.
.
.
.
1
1
−
a
[
(
1
+
a
)
+
(
1
−
a
2
)
x
+
(
1
−
a
3
)
x
2
+
(
1
−
a
4
)
x
3
.
.
.
.
.
.
.
.
.
.
]
1
1
−
a
[
1
+
x
+
x
2
+
x
3
+
.
.
.
.
.
.
.
.
.
−
(
a
+
a
2
x
+
a
3
x
2
+
a
4
x
3
+
.
.
.
.
)
]
1
1
−
a
[
1
1
−
x
−
a
1
−
a
x
]
1
1
−
a
[
1
−
a
x
−
a
+
a
x
(
1
−
x
)
(
1
−
a
x
)
]
1
(
1
−
x
)
(
1
−
a
x
)
Then,
Option
A
is correct answer.
The sum of the series
n
∑
r
=
1
(
−
1
)
r
−
1
.
n
C
r
(
a
−
r
)
is
Report Question
0%
a
0%
0
0%
n
.2
n
−
1
+
a
0%
None of these
Explanation
∑
n
r
=
1
(
−
1
)
r
−
1
n
c
r
(
a
−
r
)
P
u
t
n
=
1
⇒
n
c
1
(
a
−
1
)
−
n
c
2
(
a
−
2
)
+
n
c
3
(
a
−
1
)
+
.
.
.
.
.
.
.
(
−
1
)
n
−
1
n
c
n
(
a
−
n
)
⇒
a
[
n
c
1
−
n
c
2
+
n
c
3
−
n
c
4
+
n
c
5
+
.
.
.
.
.
.
.
−
n
c
n
]
−
n
c
1
+
2
n
c
2
−
3
n
c
3
+
4
n
c
4
.
.
.
.
.
.
.
.
.
=
a
[
n
c
1
−
n
c
2
+
n
c
3
−
n
c
4
+
n
c
5
+
.
.
.
.
.
.
]
−
[
n
c
1
−
2
n
c
2
+
3
n
c
3
+
4
n
c
4
+
.
.
.
.
.
.
.
]
(
1
−
x
)
n
=
n
c
0
−
n
c
1
x
+
n
c
2
x
2
−
n
c
3
x
3
.
.
.
.
.
.
.
[
A
c
c
o
r
d
i
n
g
t
o
t
h
i
s
f
o
r
m
u
l
a
]
⇒
A
d
d
&
s
u
b
t
r
a
c
t
n
c
0
i
n
a
h
a
v
e
e
q
u
a
t
i
o
n
&
t
a
k
i
n
g
o
u
t
c
o
m
m
o
n
(
−
1
)
a
l
s
o
.
⇒
a
[
−
(
n
c
0
−
n
c
1
+
n
c
2
−
n
c
3
+
.
.
.
.
.
.
)
+
n
c
o
]
−
[
n
c
1
−
2
n
c
2
+
3
n
c
3
−
4
n
c
4
]
P
u
t
x
=
1
(
1
−
1
)
n
=
n
c
0
−
n
c
1
+
n
c
2
−
n
c
3
=
0
S
o
,
a
[
−
0
+
n
c
0
]
−
[
n
c
1
−
2
n
c
2
+
3
n
c
3
−
4
n
c
4
+
.
.
.
.
.
.
.
.
.
]
→
(
i
)
(
1
+
n
)
n
=
n
c
0
+
n
c
1
x
+
n
c
2
x
2
+
n
c
3
x
3
+
n
c
4
x
4
+
.
.
.
.
.
.
.
[
←
f
o
r
m
u
l
a
]
D
i
f
f
.
w
.
r
.
t
.
n
,
w
e
g
e
t
n
(
1
+
x
)
n
−
1
=
n
c
1
+
2
x
n
c
2
+
3
x
n
c
3
+
4
x
n
c
4
+
.
.
.
.
P
u
t
x
=
−
1
S
o
,
x
(
1
−
x
)
n
−
1
=
n
c
1
−
2
x
n
c
2
+
3
x
2
n
c
3
+
4
n
c
4
+
.
.
.
.
.
.
.
.
.
0
=
n
c
1
−
2
n
c
2
+
3
n
c
3
−
4
n
c
4
So, put the value in equation in (1)
=
a
[
n
c
0
]
−
[
0
]
⇒
a
(
1
)
=
a
∑
n
r
=
1
(
−
1
)
r
−
1
n
c
r
(
a
−
r
)
=
a
1.3.4
+
2.5.8
+
3.6.9
+
upto
n
terms is equal to ________ .
Report Question
0%
n
(
n
+
1
)
(
n
+
2
)
6
0%
n
(
n
+
1
)
(
3
n
2
+
23
n
+
46
)
12
0%
n
(
27
n
3
+
90
n
2
+
45
n
−
5
)
4
0%
n
(
n
+
1
)
(
2
n
+
1
)
6
0%
N
o
n
e
o
f
t
h
e
s
e
The sum of the series
1
+
2.2
+
3.2
2
+
4.2
3
+
5.2
4
+
.
.
.
+
1000.2
999
is
Report Question
0%
999.2
999
−
1
0%
999.2
1000
−
1
0%
999.2
1000
+
1
0%
999.2
999
+
1
Explanation
S
=
1
+
2.2
+
3.2
2
+
4.2
3
+
5.2
4
+
.
.
.
.
.
+
1000.2
999
__ (1)
Multiply with '2' on both sides
2
S
=
1.2
+
2.2
2
+
3.2
2
+
4.2
4
+
5.2
5
+
.
.
.
+
999.2
999
+
1000.2
1000
__ (2)
subtracting equation (1) & (2)
(
S
−
2
S
)
=
[
1
+
(
2.2
−
1.2
)
+
(
3.2
2
−
2.2
2
)
+
(
4.2
3
−
3.2
3
)
+
(
5.2
4
−
4.2
4
)
+
...
+
(
1000.2
999
−
999.2
999
)
−
1000.2
1000
(
S
−
2
S
)
=
1
+
1.2
+
1.2
2
+
1.2
3
+
1.2
4
+
.
.
.
.
+
1.2
999
−
1000.2
1000
−
S
=
1
+
(
2
+
2
2
+
2
3
+
2
4
+
.
.
.
+
2
999
)
−
1000.2
1000
[
∵
a
+
a
r
+
a
r
2
+
.
.
.
+
a
r
n
−
1
=
a
.
(
r
n
−
1
)
r
−
1
]
a
=
2
,
r
=
2
,
n
=
999
−
S
=
1
+
2.
(
2
999
−
1
)
2
−
1
−
1000.2
1000
−
S
=
1
+
2
1000
−
2
−
1000.2
1000
−
S
=
−
1
+
2
1000
(
1
−
1000
)
−
S
=
−
1
−
999.2
1000
S
=
999.2
1000
+
1
n
∑
r
=
1
r
−
1
∑
p
=
0
n
C
r
⋅
r
C
p
⋅
2
p
is equal to?
Report Question
0%
4
n
−
3
n
+
1
0%
4
n
−
3
n
−
1
0%
4
n
−
3
n
+
2
0%
4
n
−
3
n
If
|
x
|
<
1
, then the sum of series
1
+
2
x
+
3
x
2
+
4
x
3
+
.
.
.
.
.
.
.
.
.
∞
will be
Report Question
0%
1
1
−
x
0%
1
1
+
x
0%
1
(
1
+
x
)
2
0%
1
(
1
−
x
)
2
Explanation
Let
S
=
1
+
2
x
+
3
x
2
+
.
.
.
.
.
.
∞
…
(
1
)
multiply by
x
on both sides
x
S
=
x
+
2
x
2
+
3
x
3
+
.
.
.
.
.
∞
…
(
2
)
Subtract
(
1
)
from
(
2
)
S
=
1
+
2
x
+
3
x
2
+
.
.
.
.
.
.
∞
x
S
=
x
+
2
x
2
+
3
x
3
+
.
.
.
.
.
∞
- - - -
--------------------------------------------------------
S
(
1
−
x
)
=
1
+
x
+
x
2
+
x
3
+
.
.
.
.
.
∞
S
(
1
−
x
)
=
1
1
−
x
[sum of infinte G.P
S
∞
=
1
1
−
r
]
S
=
1
(
1
−
x
)
2
1
2.5
+
1
5.8
+
1
8.11
+
.
.
n
terms
=
Report Question
0%
3
n
2
(
3
n
+
2
)
0%
3
n
3
n
+
2
0%
n
2
(
3
n
+
2
)
0%
n
3
n
+
2
Explanation
1
2.5
+
1
5.8
+
1
8.11
+
.
.
.
.
.
+
1
[
2
+
3
(
n
−
1
)
]
.
[
5
+
3
(
n
−
1
)
]
=
n
∑
n
=
1
1
[
2
+
3
n
−
3
]
.
[
5
+
3
n
−
3
]
=
n
∑
n
=
1
1
(
3
n
−
1
)
.
(
3
n
+
2
)
=
1
3
∑
[
3
n
+
2
−
(
3
n
−
1
)
]
(
3
n
−
1
)
(
3
n
+
2
)
=
1
3
∑
1
3
n
−
1
−
1
3
n
+
2
=
1
3
[
1
2
−
1
5
+
1
5
−
1
8
+
1
8
−
1
11
+
.
.
.
+
1
3
n
−
4
−
1
3
n
−
1
+
1
3
n
−
1
−
1
3
n
+
2
]
=
1
3
[
1
2
−
1
3
n
+
2
]
=
3
n
3
(
3
n
+
2
)
.2
=
n
2
(
3
n
+
2
)
The sum of the series
5
13
+
55
13
2
+
555
13
3
+
.
.
.
.
.
.
.
.
.
∞
is
Report Question
0%
65
36
0%
65
32
0%
25
36
0%
none of these
Explanation
S
=
5
13
+
55
13
2
+
555
13
3
+
5555
13
4
+
.
.
.
.
.
(
1
)
Now multiply by
1
13
, we get
1
13
s
=
5
13
2
+
55
13
3
+
555
13
4
+
.
.
.
.
.
(
2
)
Subtracting
(
2
)
from
(
1
)
s
−
1
13
s
[
5
13
+
55
13
2
+
555
13
3
+
.
.
.
.
.
]
−
[
5
13
2
+
55
13
3
+
.
.
…
]
12
s
13
=
5
13
+
(
55
13
2
+
5
13
2
)
+
(
555
13
3
+
55
13
3
)
+
.
.
.
.
.
12
s
13
=
5
13
+
50
13
2
+
500
13
3
+
.
.
.
.
=
5
13
1
−
10
13
12
s
13
=
5
3
⇒
s
=
65
36
.
Sum to n terms the following series :
Report Question
0%
5 + 11 + 19 + 29 + 41 + .......
0%
3 + 7 + 14 + 24 + 37 +......
0%
6 + 9 + 16 + 27 + 42 + .....
0%
5 + 7 + 13 + 31 + 85 + ....
2
n
+
1
∑
k
=
1
(
−
1
)
k
−
1
k
2
=
Report Question
0%
(
n
+
1
)
(
2
n
+
1
)
0%
(
n
+
1
)
(
2
n
−
1
)
0%
(
n
−
1
)
(
2
n
−
1
)
0%
(
n
−
1
)
(
2
n
+
1
)
Explanation
2
n
+
1
∑
k
=
1
(
−
1
)
k
−
1
k
2
=
1
2
−
2
2
+
3
2
−
4
2
+
5
2
.
.
.
.
.
.
−
(
2
n
)
2
+
(
2
n
+
1
)
2
=
1
2
+
3
2
+
5
2
+
.
.
.
.
.
+
(
2
n
+
1
)
2
−
[
2
2
+
4
2
+
.
.
.
+
(
2
n
)
2
]
=
1
2
+
2
2
+
3
2
+
.
.
.
.
.
+
(
2
n
+
1
)
2
−
2
[
2
2
+
4
2
+
.
.
.
.
(
2
n
)
2
]
Adding and subtract even terms to complete the series.
Using identify
1
2
+
2
2
+
.
.
.
.
+
n
2
=
n
(
n
+
1
)
(
2
n
+
1
)
6
The addition will become
=
1
2
+
2
2
+
3
2
+
.
.
.
.
+
(
2
n
+
1
)
−
2
(
2
2
)
[
1
2
+
2
2
+
.
.
.
.
.
+
n
2
]
=
(
2
n
+
1
)
(
2
n
+
2
)
(
4
n
+
3
)
6
−
8
n
(
n
+
1
)
(
2
n
+
1
)
6
=
2
(
2
n
+
1
)
(
n
+
1
)
(
4
n
+
3
)
6
−
8
n
(
n
+
1
)
(
2
n
+
1
)
6
=
2
6
(
n
+
1
)
(
2
n
+
1
)
[
4
n
+
3
−
4
n
]
=
1
3
(
n
+
1
)
(
2
n
+
1
)
(
3
)
=
(
n
+
1
)
(
2
n
+
1
)
If
S
=
1
+
1
1
+
2
+
1
1
+
2
+
3
+
1
1
+
2
+
3
+
4
.
.
.
.
.
.
,
then
Report Question
0%
S
n
=
2
n
n
+
1
0%
S
n
=
2
n
n
−
1
0%
S
∞
=
2
0%
S
∞
=
1
Explanation
S
n
=
1
1
+
1
1
+
2
+
1
1
+
2
+
3
+
1
1
+
2
+
3
+
4
+
.
.
.
.
.
.
T
n
=
1
1
+
2
+
3
+
.
.
.
n
=
1
n
(
n
+
1
)
2
T
n
=
2
n
(
n
+
1
)
=
2
[
(
n
+
1
)
−
n
n
(
n
+
1
)
]
=
2
[
1
n
−
1
n
+
1
]
T
1
=
2
[
1
1
−
1
2
]
T
2
=
2
[
1
2
−
1
3
]
.
.
.
.
T
n
=
2
[
1
n
−
1
n
+
1
]
Add the above terms
S
n
=
2
[
1
−
1
n
+
1
]
=
2
[
n
+
1
−
1
n
+
1
]
=
2
n
n
+
1
S
∞
=
lim
n
→
0
2
n
n
+
1
=
lim
n
→
∞
=
2
1
+
1
n
S
∞
=
2
.
If
S
=
tan
−
1
(
1
n
2
+
n
+
1
)
+
tan
−
1
(
1
n
2
+
3
n
+
3
)
+
…
+
tan
−
1
(
1
1
+
(
n
+
19
)
(
n
+
20
)
)
then
tan
S
is equal to
Report Question
0%
20
401
+
20
n
0%
n
n
2
+
20
n
+
1
0%
20
n
2
+
20
n
+
1
0%
n
401
+
20
n
Explanation
S
=
tan
−
1
(
1
n
2
+
n
+
1
)
+
tan
−
1
(
1
n
2
+
3
n
+
3
)
+
.
.
.
.
+
tan
−
1
(
1
1
+
(
n
+
19
)
(
n
+
20
)
)
S
=
tan
−
1
(
n
+
1
−
n
1
+
n
(
n
+
1
)
)
+
tan
−
1
[
(
n
+
2
)
−
(
n
−
1
)
1
+
(
n
+
1
)
(
n
+
2
)
]
+
.
.
.
.
.
+
tan
−
1
[
(
n
+
20
)
−
(
n
+
19
)
1
+
(
n
+
19
)
(
n
+
20
)
]
=
[
tan
−
1
(
n
+
1
)
−
tan
−
1
(
n
)
]
+
[
tan
−
1
(
n
+
2
)
−
tan
−
1
(
n
+
1
)
]
+
.
.
.
.
.
.
+
[
tan
−
1
(
n
+
20
)
−
tan
−
1
(
n
+
19
)
]
=
tan
−
1
(
n
+
20
)
−
t
a
n
−
1
(
n
)
=
tan
−
1
(
20
1
+
n
2
+
20
n
)
t
a
n
s
=
20
n
2
+
20
n
+
1
H
e
n
c
e
,
t
h
e
o
p
t
i
o
n
C
i
s
c
o
r
r
e
c
t
a
n
s
w
e
r
.
7, 11, 23, 51, 103 ?
Report Question
0%
186
0%
188
0%
185
0%
187
0%
none of these
Explanation
7
,
11
,
23
,
51
,
103
,
?
7
+
4
×
1
=
11
11
+
4
×
3
=
23
←
difference of 2
23
+
4
×
7
=
51
←
difference of 4
51
+
4
×
13
=
103
←
difference of 6
103
+
4
×
21
=
187
←
difference will be of 8
The value of the expression
n
∑
r
=
0
(
−
1
)
r
(
n
C
r
r
+
3
C
r
)
is
Report Question
0%
n
(
n
+
1
)
2
0%
n
+
3
3
0%
3
n
+
3
0%
n
+
2
2
S
=
tan
−
1
(
1
n
2
+
n
+
1
)
+
tan
−
1
(
1
n
2
+
3
n
+
3
)
+
.
.
.
.
.
+
tan
−
1
(
1
1
+
(
n
+
19
)
(
n
+
20
)
)
, then
tan
S
is equal to?
Report Question
0%
20
401
+
20
n
0%
n
n
2
+
20
n
+
1
0%
20
n
2
+
20
n
+
1
0%
n
401
+
20
n
Explanation
S
=
19
∑
r
=
0
t
a
n
−
1
(
1
1
+
(
n
+
r
)
(
n
+
r
+
1
)
)
=
t
a
n
−
1
(
(
n
+
r
+
1
)
−
(
n
+
r
)
1
+
(
n
+
r
)
(
n
+
r
+
1
)
)
⇒
S
=
19
∑
r
=
0
(
t
a
n
−
1
(
n
+
r
+
1
)
−
t
a
n
−
1
(
n
+
r
)
)
=
t
a
n
−
1
(
n
+
1
)
−
t
a
n
−
1
(
n
)
=
t
a
n
−
1
(
n
+
20
)
−
t
a
n
−
1
(
n
+
19
)
+
t
a
n
−
1
(
n
+
2
)
−
t
a
n
−
1
(
n
+
1
)
t
a
n
−
1
(
n
+
20
)
−
t
a
n
−
1
(
n
+
19
)
⇒
t
a
n
S
=
(
n
+
20
)
−
(
n
)
1
+
n
(
n
+
20
)
=
20
n
2
+
20
n
+
1
⇒
(
C
)
13, 16, 22, 33, 51 ?
Report Question
0%
89
0%
78
0%
102
0%
69
0%
none of these
Explanation
REF.Image
13
,
16
,
22
,
33
,
51
,
?
Evaluate:-
If
n
∑
r
−
0
{
n
C
r
−
1
n
C
r
+
n
C
r
−
1
}
3
=
25
24
Report Question
0%
3
0%
6
0%
4
0%
5
655, 439, 314, 250, 223 ?
Report Question
0%
215
0%
210
0%
195
0%
190
0%
none of these
Explanation
Solution
→
Given terms are 655,439,314,250,223,?
As
a
1
=
655
a
2
=
439
a
3
=
314
a
4
=
250
a
5
=
223
a
6
=
?
So,
a
1
−
a
2
=
655
−
439
=
216
=
6
3
a
3
−
a
2
=
439
−
314
=
125
=
5
3
a
3
−
a
4
=
−
250
+
314
=
64
=
4
3
a
4
−
a
5
=
−
223
+
250
=
27
=
3
3
So, we can observe difference b/w terms are cubes
of 6,5,4,3 so, is similar way
a
5
−
a
6
=
2
3
223
−
a
6
=
8
a
6
=
215
So, we find next term is 215
If the expansion of
(
x
+
a
)
n
if the sum of odd terms be P & sum of even terms be Q, prove that
Report Question
0%
P
2
−
Q
2
=
(
x
2
−
a
2
)
n
0%
4
P
Q
=
(
x
+
a
)
2
n
−
(
x
−
a
)
2
n
0%
P
2
−
Q
2
=
(
x
2
+
a
2
)
n
0%
N
o
n
e
o
f
t
h
e
s
e
Sum of the series
S
=
1
2
−
2
2
+
3
2
−
4
2
+
.
.
.
.
.
−
2000
2
+
2003
2
is
Report Question
0%
2007006
0%
1005004
0%
2000506
0%
N
o
n
e
Find the next term
210
,
209
,
205
,
196
,
180
,
?
Report Question
0%
138
0%
77
0%
155
0%
327
Explanation
T
1
=
210
,
T
2
=
209
,
T
3
=
205
,
T
4
=
196
,
T
5
=
180
,
T
6
=
?
T
1
−
T
2
=
1
2
T
2
−
T
3
=
2
2
T
3
−
T
4
=
3
2
T
4
−
T
5
=
4
2
T
5
−
T
6
=
5
2
⟹
T
6
=
T
5
−
5
2
=
180
−
25
=
155
Determine the next term
20
,
24
,
33
,
49
,
74
,
110
,
?
Report Question
0%
133
0%
147
0%
159
0%
163
0%
171
Explanation
20
,
24
,
33
,
49
,
74
,
110
,
.
.
.
Let the number be
x
difference
24
−
20
=
4
,
33
−
24
=
9
,
49
−
33
=
16
,
74
−
49
=
25
,
110
−
74
=
36
,
x
−
110
=
49
⟹
24
−
20
=
2
2
,
33
−
24
=
3
2
,
49
−
33
=
4
2
,
74
−
49
=
5
2
110
−
74
=
6
2
,
x
−
110
=
7
2
So,
110
+
49
=
159
ANSWER IS C
Find the next term of the series
1728
,
2744
,
4096
,
5832
,
8000
,
10648
,
?
Report Question
0%
2167
0%
13824
0%
15625
0%
9261
0%
17576
Explanation
1728
,
2744
,
4096
,
5832
,
8000
,
10648.....
12
3
→
1728
14
3
→
2744
16
3
→
4096
18
3
→
5832
20
3
→
8000
22
3
→
10648
24
3
→
13824
ANSWER IS B
4
,
6
,
12
,
30
,
90
,
315
,
?
Report Question
0%
945
0%
1102
0%
1260
0%
1417.5
0%
N
o
n
e
o
f
t
h
e
s
e
Explanation
=
315
×
4
=
1260.
Hence, the answer is
1260.
Find next term
462
,
552
,
650
,
756
,
870
,
992
,
?
Report Question
0%
1040
0%
1122
0%
1132
0%
105
Find the missing number from the given alternatives.
Report Question
0%
39
,
116
0%
52
,
156
0%
30
,
117
0%
31
,
116
Explanation
13
39
26
78
?
?
The pattern satisfies
x
3
x
Therefore correct option is
52
156
8
,
31
,
122
,
485
,
1936
,
7739
,
?
Report Question
0%
30950
0%
46430
0%
34650
0%
42850
0%
38540
The sum of infinite series
|
1
2
6
4
|
+
|
1
2
2
2
4
|
+
|
1
4
2
2
3
4
|
+
.
.
.
.
.
.
.
.
.
Report Question
0%
−
10
0%
0
0%
10
0%
∞
If
π
4
−
1
+
1
3
−
1
5
+
1
7
−
1
9
+
1
11
−
1
13
+
=
0
then value of
1
1
×
3
+
1
5
×
7
+
1
9
×
11
+
1
13
×
15
+
.
.
is
Report Question
0%
π
8
0%
π
6
0%
π
4
0%
π
34
Let
t
r
=
r
1
+
r
2
+
r
4
then,
lim
n
→
∞
n
∑
r
=
1
t
r
equals
Report Question
0%
1
4
0%
1
0%
1
2
0%
None of these
Explanation
t
r
=
r
1
+
r
2
+
r
4
=
r
r
4
+
2
r
2
+
1
−
r
2
=
r
(
r
2
+
1
)
2
−
r
2
=
r
(
r
2
+
1
−
r
)
(
r
2
+
1
+
r
)
=
1
2
2
r
(
r
2
+
1
−
r
)
(
r
2
+
r
+
1
)
=
1
2
(
r
2
+
r
+
1
)
−
(
r
2
−
r
+
1
)
(
r
2
−
r
+
1
)
(
r
2
+
r
+
1
)
=
1
2
[
1
r
2
−
r
+
1
−
1
r
2
+
r
+
1
]
=
1
2
[
1
r
2
−
r
+
1
−
1
(
r
+
1
)
2
−
(
r
+
1
)
+
1
]
n
∑
r
=
1
t
r
=
1
2
[
1
1
−
1
+
1
−
1
(
n
+
1
)
2
−
(
n
+
1
)
+
1
]
=
1
2
[
1
−
1
n
2
+
n
+
1
]
=
1
2
[
n
2
+
n
+
1
−
1
n
2
+
n
+
1
]
∑
t
n
=
1
2
[
1
+
1
n
1
+
1
n
+
1
n
2
]
l
t
n
→
∞
∑
t
r
=
1
2
[
1
+
0
1
+
0
+
0
]
=
1
2
.
The sum to
n
terms of the series
3
1
2
+
5
1
2
+
2
2
+
7
1
2
+
2
2
+
3
2
+
.
.
.
.
.
.
.
.
.
is
Report Question
0%
6
n
n
+
1
0%
9
n
n
+
1
0%
12
n
n
+
1
0%
3
n
n
+
1
Explanation
L
e
t
T
r
=
2
r
+
1
1
2
+
2
2
+
r
2
=
(
2
r
+
1
)
r
(
r
+
1
)
(
2
r
+
1
)
=
6
r
(
2
+
1
)
∑
T
r
=
∑
6
r
(
r
+
1
)
=
6
∑
n
r
=
1
(
r
+
1
)
−
r
r
(
r
+
1
)
=
6
∑
n
r
=
1
[
1
r
−
1
r
+
1
]
=
6
[
(
1
−
1
2
)
+
(
1
2
−
1
3
)
+
.
.
.
.
.
.
.
+
(
1
n
−
1
n
+
1
)
]
=
6
[
1
−
1
n
+
1
]
=
6
n
n
+
1
H
e
n
c
e
,
o
p
t
i
o
n
A
i
s
t
h
e
c
o
r
r
e
c
t
a
n
s
w
e
r
.
499
,
622
,
868
,
1237
,
1729
,
2344
,
?
Report Question
0%
3205
0%
3082
0%
2959
0%
3462
0%
2876
Explanation
499
,
622
,
868
,
1237
,
1729
,
2344
,
______
499
+
1
×
123
=
499
+
123
=
622
622
+
2
×
123
=
622
+
146
=
868
868
+
3
×
123
=
868
+
369
=
1237
1237
+
4
×
123
=
1237
+
492
=
1729
1729
+
5
×
123
=
1729
+
615
=
2344
2344
+
6
×
123
=
2344
+
738
=
3082
.
The sum
1
1
+
1
2
+
1
4
+
2
1
+
2
2
+
2
4
+
3
1
+
3
2
+
3
4
+
.
.
.
+
99
1
+
99
2
+
99
4
lies between
Report Question
0%
0.46
and
0.47
0%
0.52
and
1.0
0%
0.48
and
0.49
0%
0.49
and
0.50
Explanation
We have,
1
1
+
1
2
+
1
4
+
2
1
+
2
2
+
2
4
+
3
1
3
+
2
2
+
3
4
+
.
.
.
.
.
.
.
.
.
+
99
1
+
99
2
+
99
4
=
99
∑
n
=
1
x
1
+
x
2
+
x
4
=
99
∑
n
=
1
x
(
x
2
+
1
)
2
−
x
2
=
99
∑
n
=
1
(
x
2
+
x
+
1
)
−
(
x
2
−
x
+
1
)
2
(
x
2
+
x
+
1
)
(
x
2
−
x
+
1
)
=
1
2
99
∑
n
=
1
(
1
x
2
−
x
+
1
−
1
x
2
+
x
+
1
)
=
1
2
(
1
1
−
1
3
+
1
3
−
1
7
+
1
7
+
.
.
.
.
.
.
.
.
.
+
1
99
2
+
99
+
1
)
=
1
2
(
1
−
1
9901
)
=
4950
9901
≈
1
2
H
e
n
c
e
,
o
p
t
i
o
n
D
i
s
c
o
r
r
e
c
t
a
n
s
w
e
r
.
Find :
12
,
15
,
21
,
24
,
30
,
33
,
?
,
?
Report Question
0%
39
,
42
0%
37
,
42
0%
38
,
47
0%
39
,
51
Explanation
12
,
15
,
21
,
24
,
30
,
33
,
?
,
?
12
12
+
3
=
15
15
+
3
×
2
=
21
21
+
3
=
24
24
+
3
×
2
=
30
30
+
3
=
33
33
+
3
×
2
=
39
39
+
3
=
42
Find the missing number :
Report Question
0%
125
0%
90
0%
105
0%
225
Explanation
Step 1: Observe the numbers opposite to each other
We can see that the number opposite to 315 is 21. Also,
315
21
=
15
Hence, we can say the missing number has at least one factor of 15
We see the missing number is opposite to 15
Also,
15
×
15
=
225
Step 2: Verification
Let the missing number be x.
ATQ,
⇒
21
+
15
+
x
=
261
⇒
x
=
261
−
21
−
15
x
=
225
Hence, it is verified that x=225
Thus, the missing number is D 225.
23
,
29
,
47
,
75
,
?
Report Question
0%
87
0%
93
0%
110
0%
117
Explanation
2
3
,
2
⏟
9
,
4
⏟
7
,
7
⏟
5
,
?
⏟
=
75
+
35
6
18
28
35
(
2
×
3
)
↑
(
2
×
9
)
↑
(
4
×
7
)
↑
(
7
×
5
)
↑
=
110
The value of
1
6.10
+
1
10.14
+
1
14.18
+
.
.
.
.
∞
equals to
Report Question
0%
1
(
24
)
2
0%
1
6
0%
1
24
0%
1
(
24
)
3
Explanation
S
=
1
6
⋅
10
,
1
10
⋅
14
+
1
1419
+
.
.
.
.
.
∞
S
=
1
4
(
1
6
−
1
10
)
+
1
4
(
1
10
−
1
14
)
+
1
4
(
1
14
−
1
18
)
+
.
.
.
.
.
∞
Now take
1
4
common
S
=
1
4
[
1
6
−
1
16
+
1
10
−
1
1
/
4
+
1
14
−
1
18
+
1
,
−
∞
]
S
=
1
4
×
1
6
=
1
24
option
c
is correct
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
1
Not Answered
49
Not Visited
Correct : 0
Incorrect : 0
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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