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CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 1 - MCQExams.com
CBSE
Class 10 Maths
Arithmetic Progressions
Quiz 1
In an arithmetic series,
a
1
=
−
14
and
a
5
=
50
. Find the sum of the first 5 terms.
Report Question
0%
60
0%
70
0%
80
0%
90
Explanation
First term = a = -14
last term, l = 50
a
n
=
50
n = 5
S
n
=
n
2
[
a
+
l
]
S
5
=
5
2
[
−
14
+
50
]
S
5
=
2.5
[
36
]
S
5
=
90
−
20
,
−
16
,
−
12
,
−
8.....
is the sequence and each term is increased by
4
. Then which of the following could not be a term in the sequence?
Report Question
0%
762
0%
200
0%
440
0%
0
The sum of first
20
natural numbers is
Report Question
0%
55
0%
210
0%
110
0%
215
Explanation
The sum of first '
n
' natural numbers is given by
n
(
n
+
1
)
2
.
So, the sum of first
20
natural numbers is
20
(
20
+
1
)
2
.
=
20
(
21
)
2
=
210
.
Find first term 'a' and common difference 'd' for the following AP.
√
2
,
√
8
,
√
18
,
√
32
,.....
Report Question
0%
a
=
√
2
,
d
=
√
4
0%
a
=
√
2
,
d
=
√
2
0%
Its not an AP
0%
None of these
Explanation
The sequence
√
2
,
√
8
,
√
18
,
√
32
,
.
.
.
can be written as
√
2
,
2
√
2
,
3
√
2
,
4
√
2
,
.
.
.
Clearly, it is an AP with first term
a
=
√
2
and common difference
d
=
√
2
.
The sum of the first
1000
positive integer is
Report Question
0%
500500
0%
650650
0%
300300
0%
700700
Explanation
Sum of '
n
' natural numbers
=
n
(
n
+
1
)
2
.
So, sum of first
1000
natural numbers
=
1000
(
1000
+
1
)
2
=
500
(
1001
)
=
500500
.
Hence, option A is correct.
State the following statement is true or false:
Progression means increment of quantity in a particular pattern.
Report Question
0%
True
0%
False
Explanation
Progression means increment of quantity in a particular pattern.
For example:
2
,
4
,
6
,
8
,
.
.
.
.
and
3
,
6
,
12
,
24
,
.
.
.
.
Which term of the sequence
−
7
,
−
2
,
3
,
8
,
.
.
.
.
.
.
.
.
.
is
73
?
Report Question
0%
17
0%
18
0%
16
0%
none
Explanation
Here first term
a
=
−
7
, common difference
d
=
−
2
−
(
−
7
)
=
5
and last term
l
=
73
Since,
l
=
a
+
(
n
−
1
)
d
⇒
73
=
−
7
+
(
n
−
1
)
5
⇒
80
=
(
n
−
1
)
5
⇒
16
=
n
−
1
⇒
n
=
17
∴
Option A is correct.
Find the
11
th term of the sequence
5
,
2
,
−
1
,
−
4
,
.
.
.
.
.
.
Report Question
0%
−
25
0%
−
28
0%
−
22
0%
None
Explanation
a
1
=
5
;
d
=
a
2
−
a
1
=
2
−
5
=
−
3
∴
a
11
=
a
+
(
n
−
1
)
d
=
5
+
10
(
−
3
)
⇒
a
11
=
−
25
The nth term of the sequence
2
,
4
,
6
,
8....
is
Report Question
0%
n
0%
n
−
1
0%
2
n
0%
2
n
−
4
Explanation
Clearly, the difference of successive terms of above sequence is constant which is 2
So given sequence is in AP with first term 2 and common difference
2
Hence general term is,
a
n
=
a
+
(
n
−
1
)
d
=
2
+
(
n
−
1
)
2
=
2
n
Hence option 'C' is correct choice
Find the A.P. whose sum to
n
terms is
2
n
2
+
n
Report Question
0%
The required A.P. is
2
,
6
,
10
,
14
,
.
.
.
0%
The required A.P. is
3
,
7
,
11
,
15
,
.
.
.
0%
The required A.P. is
4
,
8
,
12
,
16
,
.
.
.
0%
The required A.P. is
5
,
10
,
15
,
20
,
.
.
.
Explanation
Given,
S
n
=
2
n
2
+
n
Now,
a
1
=
S
1
=
2
(
1
)
2
+
1
=
3
a
2
=
S
2
−
S
1
=
2
(
2
)
2
+
2
−
3
=
7
a
3
=
S
3
−
S
2
=
[
2
(
3
)
2
+
3
]
−
[
2
(
2
)
2
+
2
]
=
21
−
10
=
11
a
4
=
S
4
−
S
3
=
[
2
(
4
)
2
+
4
]
−
[
2
(
3
)
2
+
3
]
=
36
−
21
=
15
and so on.
Therefore the required A.P. is
3
,
7
,
11
,
15...
If the common difference of an AP is
−
6
,
then what is
a
16
−
a
12
?
Report Question
0%
−
24
0%
24
0%
−
30
0%
30
Explanation
Given,
d
=
−
6
a
16
=
a
+
15
d
a
12
=
a
+
11
d
a
16
−
a
12
=
(
a
+
15
d
)
−
(
a
+
11
d
)
=
4
d
∴
a
16
−
a
12
=
4
×
−
6
=
−
24
If 100 times the
100
t
h
term of an AP with non zero common difference equals the 50 times its
50
t
h
term, then the
150
t
h
term of this AP is:
Report Question
0%
−
150
0%
150
times its
50
t
h
term
0%
150
0%
0
Explanation
Let say
a
be the first term and
d
be the common difference of the AP.
Given:
100
(
a
+
99
d
)
=
50
(
a
+
49
d
)
[using
n
t
h
term
a
n
=
a
+
(
n
−
1
)
d
]
⇒
2
a
+
198
d
=
a
+
49
d
⇒
a
+
149
d
=
0
Also,
T
150
=
a
+
149
d
=
0
Hence, the option 'D' is correct.
Sum of first
15
terms of
2
+
5
+
8
+
.
.
.
is?
Report Question
0%
44
0%
42
0%
345
0%
386
Explanation
Difference between terms
=
5
−
2
=
8
−
5
=
3
We can observe that difference between two consecutive terms is equal, hence the terms are in A.P.
Sum of fist n terms in an A.P. is given by the formula:
S
n
=
n
2
×
[
2
a
+
(
n
−
1
)
d
]
Given
n
=
15
,
a
=
2
,
d
=
5
−
2
=
3
,
S
n
=
15
2
×
[
2
×
2
+
(
15
−
1
)
3
]
S
n
=
15
2
×
46
S
n
=
345
Check whether the following form an AP
√
3
,
√
12
,
√
27
,
√
48
, ...
Report Question
0%
True
0%
False
Explanation
We can write the given series as
√
3
,
2
√
3
,
3
√
3
,
4
√
3
Compare the given series with
a
1
,
a
2
,
a
3
.
.
.
.
.
.
we get
a
2
−
a
1
=
2
√
3
−
√
3
=
√
3
a
3
−
a
2
=
3
√
3
−
2
√
3
=
√
3
and so on.
Since common difference is same and equal to
√
3
, therefore given series is AP.
Which of the following is not an A.P.?
Report Question
0%
13
,
8
,
3
,
−
2
,
−
7
,
12
0%
10.8
,
11.2
,
11.6
,
12
,
12.4
0%
8
1
7
,
18
2
7
,
28
3
7
,
48
4
7
,
58
5
7
0%
8
3
23
,
11
6
23
,
14
9
23
,
17
12
23
Explanation
(
A
)
The last term of this sequence is not in AP
(
B
)
The sequence is in AP with common difference
0.4
(
C
)
57
7
,
128
7
,
199
7
,
340
7
,
411
7
The last two terms of this sequence are not in AP
(
D
)
187
23
,
259
23
,
331
23
,
403
23
The sequence is in AP with common difference
72
23
The list of numbers
10
,
6
,
2
,
−
2
,
⋯
is
Report Question
0%
an
A
.
P
.
with
d
=
16
0%
an
A
.
P
.
with
d
=
−
4
0%
an
A
.
P
.
with
d
=
4
0%
not an
A
.
P
.
Explanation
Given series:
10
,
6
,
2
,
−
2...
Let,
a
1
=
10
,
a
2
=
6
,
a
3
=
2
,
a
4
=
−
2
d
1
=
a
2
−
a
1
=
6
−
10
=
−
4
d
2
=
a
3
−
a
2
=
2
−
6
=
−
4
d
3
=
a
4
−
a
3
=
−
2
−
2
=
−
4
The differences are equal, hence common difference of this A.P. is
d
=
−
4
The common difference of the AP
1
P
,
1
−
P
P
,
1
−
2
P
P
, .......... is:
Report Question
0%
P
0%
-1
0%
-P
0%
1
Explanation
Let
a
1
=
1
P
,
a
2
=
1
−
P
P
,
a
3
=
1
−
2
P
P
∴
Common difference =
a
2
−
a
1
o
r
a
3
−
a
2
;
=
(
1
−
P
)
−
(
1
)
P
;
(
1
−
2
P
)
−
(
1
−
P
)
P
=
−
1
What is the sum of all natural numbers from 1 to 100?
Report Question
0%
5050
0%
50
0%
4550
0%
5150
Explanation
We know,
Sum of n natural numbers
=
n
(
n
+
1
)
2
∴
Sum from 1 to 100
=
100.101
2
=
5050
(
A
n
s
→
A
)
If the first term of an A.P. is
−
1
and common difference is
−
3
, then its
12
t
h
term is
Report Question
0%
34
0%
32
0%
−
32
0%
−
34
Explanation
Given that,
The first term of an A.P.
(
a
)
=
−
1
,
The common difference
(
d
)
=
−
3
To find out:
12
t
h
term of the A.P.
We know that,
n
t
h
term of any AP is given by:
T
n
=
a
+
(
n
−
1
)
d
Hence,
T
12
=
(
−
1
)
+
(
12
−
1
)
(
−
3
)
⇒
T
12
=
(
−
1
)
+
(
11
)
(
−
3
)
=
−
1
−
33
∴
T
12
=
−
34
Hence, the
12
t
h
term of the given A.P. is
−
34
.
Calculate
10
t
h
term of an AP:
4
,
6
,
8
,
.
.
.
.
.
.
.
.
Report Question
0%
18
0%
20
0%
22
0%
26
Explanation
Let
S
=
4
+
6
+
8
+
.
.
.
∞
The above series is an arithmetic progression with common difference of
2
.
We know that,
a
n
=
a
1
+
(
n
−
1
)
d
∴
a
10
=
4
+
(
10
−
1
)
(
2
)
=
4
+
9
(
2
)
=
18
+
4
=
22
Write the sum of first five terms of the following Arithmetic Progressions where, the common difference
d
and the first term
a
are given,
a
=
2
,
d
=
2.5
Report Question
0%
32
0%
33
0%
34
0%
35
Explanation
Here the first term
a
=
2
and common difference
d
=
2.5
.
Now
a
n
=
a
+
(
n
−
1
)
d
∴
a
1
=
a
=
2
a
2
=
a
+
d
=
2
+
2.5
=
4.5
a
3
=
a
+
2
d
=
2
+
2
×
2.5
=
7
a
4
=
a
+
3
d
=
2
+
3
×
2.5
=
9.5
a
5
=
a
+
4
d
=
2
+
4
×
2.5
=
12
∴
series in A.P. is
2
,
4.5
,
7
,
9.5
,
12
,
.
.
.
.
Then, the sum of first five terms is
=
2
+
4.5
+
7
+
9.5
+
12
=
35
Or
We can use the formula for sum of A.P
S
n
=
n
2
(
2
a
+
(
n
−
1
)
d
)
∴
S
5
=
5
2
×
[
2
(
2
)
+
(
5
−
1
)
(
2.5
)
]
=
35
A property valued at Rs. 30000 will depreciate Rs. 1380 the first year, Rs. 1340 the second year, Rs. 1300 the third year, and so on. What will the depreciation during the eight year?
Report Question
0%
Rs. 1150
0%
Rs. 1200
0%
Rs. 1100
0%
Rs. 1250
Explanation
Every year depreciation decrease by 40, so the given problem can be solved through the AP: 1380, 1340, 1300, ........
So
a
=
1380
,
d
=
40
,
n
=
8
∴
Depreciation during eight year
=
1380
+
(
8
−
1
)
×
(
−
40
)
=
R
s
.
1100
For an A.P., the seventh term is 19 and the thirteenth term isFind its twenty second term.
Report Question
0%
60
0%
54
0%
64
0%
78
Explanation
For the given A.P.,
T
7
=
19
and
T
13
=
37
Now
T
n
=
a
+
(
n
−
1
)
d
∴
T
7
=
a
+
(
7
−
1
)
d
∴
19
=
a
+
6
d
∴
a
+
6
d
=
19
............ (1)
Again,
T
13
=
a
+
(
13
−
1
)
d
∴
37
=
a
+
12
d
∴
a
+
12
d
=
37
............. (2)
Subtracting eq. (1) from eq. (2)
a
+
12
d
=
37
a
+
6
d
=
19
6
d
=
18
∴
d
=
13
Substituting
d
=
13
in
a
+
6
d
=
19
,
a
+
6
(
3
)
=
19
∴
a
+
18
=
19
∴
a
=
1
Thus, the first term of the A.P. is 1 and the common difference is 3.
Now,
T
22
=
a
+
(
22
−
1
)
d
=
1
+
(
22
−
1
)
(
3
)
=
1
+
(
21
)
(
3
)
=
1
+
63
=
64
Thus, the twenty second term of the given A.P. is 64.
In a flower bed, there are
23
rose plants in the first row,
21
in the second,
19
in the third, and so on. There are
5
rose plants in the last row. The number of rows in the flower bed are......
Report Question
0%
11
0%
41
0%
10
0%
9
Find the sum of all odd natural numbers from 1 to 150.
Report Question
0%
5625
0%
5635
0%
5645
0%
5655
Explanation
Clearly, the odd natural numbers from
1
to
150
are
1
,
3
,
5
,
.
.
.
,
149
.
This is an AP with first term
a
=
1
, common difference
d
=
2
and last term
a
n
=
l
=
149
.
Let there be
n
terms in this AP.
Then,
a
n
=
149
⇒
a
n
=
a
+
(
n
−
1
)
d
149
=
1
+
(
n
−
1
)
×
2
149
=
1
+
2
n
−
2
2
n
=
150
∴
n
=
75
∴
required sum
=
S
n
=
n
2
[
a
+
l
]
=
75
2
[
1
+
149
]
=
5625
The sum of 24 terms of the following series :
√
2
+
√
8
+
√
18
+
√
32
+
..................
Report Question
0%
300
√
2
0%
150
√
2
0%
250
√
2
0%
325
√
2
Explanation
We have
√
2
+
√
8
+
√
18
+
√
32
+
..................
We can rewrite the series as
1
√
2
+
2
√
2
+
3
√
2
+
4
√
2
+
.
.
.
.
.
.
.
.
Taking
√
2
common, we get:
√
2
[
1
+
2
+
3
+
4
+
.
.
.
.
.
.
.
upto
24
terms
]
The
24
t
h
term will be
24
.
Hence,
√
2
[
1
+
2
+
3
+
4
+
.
.
.
.
.
.
.
+
24
]
We know that, the sum of
n
terms of an AP is given by
S
n
=
n
2
(
a
+
l
)
Here,
a
=
1
,
l
=
24
,
n
=
24
∴
S
24
=
√
2
[
24
2
(
1
+
24
)
]
⇒
√
2
×
24
×
25
2
=
300
√
2
Hence, the sum of
24
terms of the given series is
300
√
2
.
Write the sum of first five terms of the following Arithmetic Progressions where, the common difference
d
and the first term
a
are given:
a
=
4
,
d
=
0
Report Question
0%
20
0%
22
0%
24
0%
26
Explanation
Here the first term
a
=
4
and common difference
d
=
0
.
Now
a
n
=
a
+
(
n
−
1
)
d
∴
a
1
=
a
=
4
a
2
=
a
+
d
=
4
+
0
=
4
a
3
=
a
+
2
d
=
4
+
2
×
0
=
4
a
4
=
a
+
3
d
=
4
+
3
×
0
=
4
a
5
=
a
+
4
d
=
4
+
4
×
0
=
4
∴
series in A.P. is
4
,
4
,
4
,
4
,
4
,
.
.
.
.
Thus, the sum of first five terms is
4
+
4
+
4
+
4
+
4
=
20
Hence, the answer is
20
.
Or
We can use the formula for sum of A.P
S
n
=
n
2
(
2
a
+
(
n
−
1
)
d
)
∴
S
5
=
5
2
×
[
2
(
4
)
+
(
5
−
1
)
(
0
)
]
=
20
Summation of
n
terms of an A.P. is
Report Question
0%
n
2
(
a
+
l
)
0%
n
2
[
2
a
+
(
n
−
1
)
d
]
0%
a
(
r
n
−
1
)
(
r
−
1
)
0%
a
(
1
−
r
n
)
(
1
−
r
)
Explanation
General equation for the sum of n terms of an AP is given by:
n
2
(
2
a
+
(
n
−
1
)
d
)
=
n
2
(
a
+
l
)
S
n
=
54
+
51
+
48
+
.
.
.
.
n
terms
=
513
. Value of
n
is
Report Question
0%
18
0%
19
0%
15
0%
None of these above
Explanation
From the given series, we have
S
n
=
513
⇒
n
2
[
2
(
54
)
+
(
n
−
1
)
(
−
3
)
]
=
513
⇒
n
(
108
−
3
n
+
3
)
=
1026
⇒
n
2
−
37
n
+
342
=
0
⇒
n
2
−
19
n
−
18
n
+
342
=
0
⇒
n
(
n
−
19
)
−
18
(
n
−
19
)
=
0
⇒
(
n
−
18
)
(
n
−
19
)
=
0
Therefore,
n
=
18
or
n
=
19
The nth term of the A.P.
3
,
7
,
11
,
15
,
.
.
.
.
is given by -
Report Question
0%
T
n
=
4
n
−
1
0%
T
n
=
3
n
+
4
0%
T
n
=
3
n
−
4
0%
T
n
=
2
n
+
1
Explanation
Given A.P. is
3
,
7
,
11
,
15
,
.
.
.
.
Here
a
=
3
,
d
=
4
We know
T
n
=
a
+
(
n
−
1
)
d
=
3
+
(
n
−
1
)
4
=
4
n
−
1
0:0:1
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30
0
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