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CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 10 - MCQExams.com
CBSE
Class 10 Maths
Arithmetic Progressions
Quiz 10
Which term of the AP :
3
,
8
,
13
,
18
,
.
.
is
78
?
Report Question
0%
t
16
0%
t
18
0%
t
14
0%
None of these
Explanation
Given :- A.P. is
3
,
8
,
13
,
18
,
.
.
.
.
To get :- which term is
78
.
solution :-
First term
=
3
=
a
1
common diff.
=
d
=
(
8
−
3
)
=
(
13
−
8
)
=
(
18
−
13
)
=
5
∴
Let the required term be n
∴
a
n
=
78
∴
a
n
=
a
1
+
(
n
−
1
)
d
⇒
78
=
3
+
(
n
−
1
)
d
⇒
78
=
3
+
(
n
−
1
)
5
⇒
78
=
3
+
5
n
−
5
⇒
78
=
−
2
+
5
n
⇒
5
n
=
78
+
2
=
80
⇒
5
n
=
80
⇒
n
=
16
∴
so the required term is
16
t
h
term.
i.e.
t
16
Find the common difference of an AP. whose first term is
100
and the sum of whose first six terms is five times the sum of the next six terms.
Report Question
0%
10
0%
−
10
0%
5
0%
−
5
Explanation
Here
a
=
100
Let difference is
d
.
⇒
a
1
+
a
2
+
a
3
+
a
4
+
a
5
+
a
6
=
5
(
a
7
+
a
8
+
a
9
+
a
10
+
a
11
+
a
12
)
So by the formula,
S
n
=
n
2
(
a
+
l
)
, where
a
&
l
are the first and last term of an AP, we have
6
(
a
1
+
a
6
2
)
=
5
×
6
(
a
7
+
a
12
2
)
⇒
a
1
+
a
6
=
5
(
a
7
+
a
12
)
⇒
a
+
a
+
5
d
=
5
(
a
+
6
d
+
a
+
11
d
)
⇒
2
a
+
5
d
=
10
a
+
85
d
⇒
80
d
=
−
8
a
⇒
d
=
−
a
10
⇒
−
100
10
⇒
−
10
If
n
t
h
term of an AP is
1
3
(
2
n
+
1
)
, then the sum of its
19
term is
Report Question
0%
131
0%
132
0%
133
0%
134
Explanation
Let
T
n
be the
n
t
h
term of an AP
⇒
T
n
=
a
+
(
n
−
1
)
d
=
n
d
+
a
−
d
On comparing with the given expression it is clear that
d
=
2
3
and
a
−
d
=
1
3
⇒
a
−
2
3
=
1
3
⇒
a
=
1
We know that s
um of first
n
terms,
S
n
=
n
2
(
2
a
+
(
n
−
1
)
d
)
, where
a
&
d
are the first term amd common difference of an AP.
∴
S
19
=
19
2
[
2
+
(
18
)
2
3
]
=
133
Hence, option
(
C
)
is correct.
Which term of AP :
3
,
15
,
27
,
39
,
.
.
will be
132
more than its
54
t
h
term?
Report Question
0%
t
65
0%
t
64
0%
t
60
0%
None of these
Explanation
Clearly, the given Sequence is an AP with first term
a
=
3
and common difference
d
=
15
−
3
=
12
.
Let
n
t
h
term be the
132
more than
54
t
h
term of the given AP.
⇒
a
n
=
a
54
+
132
⇒
a
+
(
n
−
1
)
d
=
a
+
53
d
+
132
⇒
(
n
−
1
)
×
12
=
53
×
12
+
132
⇒
n
−
1
=
53
+
11
⇒
n
=
65
Hence,
a
65
is
132
more than
a
54
The sum of the first nineteen terms of an A.P.
a
1
,
a
2
,
a
3
.......... if it is known that
a
4
+
a
8
+
a
12
+
a
16
=
224
is .......... .
Report Question
0%
1064
0%
896
0%
532
0%
448
Explanation
We know that
n
t
h
term of an AP is given as,
a
n
=
a
+
(
n
−
1
)
d
And sum of first
n
terms,
S
n
=
n
2
(
2
a
+
(
n
−
1
)
d
)
, where
a
and
d
are the first term amd common difference of an AP.
Since,
a
4
+
a
8
+
a
12
+
a
16
=
224
∴
a
+
3
d
+
a
+
7
d
+
a
+
11
d
+
a
+
15
d
=
224
⇒
4
a
+
36
d
=
224
⇒
a
+
9
d
=
56
Now
S
19
=
19
2
[
2
a
+
(
19
−
1
)
d
]
=
19
[
a
+
9
d
]
=
19
[
56
]
=
1064
If
9
t
h
and
19
t
h
terms of an AP are
35
and
75
,
then its
20
t
h
term is
Report Question
0%
78
0%
79
0%
80
0%
81
Explanation
T
9
=
35
⇒
a
+
(
9
−
1
)
d
=
35
⇒
a
+
8
d
=
35
T
19
=
75
⇒
a
+
(
19
−
1
)
d
=
75
⇒
a
+
18
d
=
75
Solving these two we get
d
=
4
,
a
=
3
Therefore,
T
20
=
a
+
(
20
−
1
)
d
=
3
+
19
×
4
=
79
The sum of all integers between
81
and
719
which
are divisible by
5
is
Report Question
0%
51800
0%
50800
0%
52800
0%
None of these
Explanation
All integers between
81
and
719
which are divisible by
5
form an AP with common difference
5
.
First term
a
=
85
, last term
l
=
715
and common difference is
d
=
5
Then, apply the formula for
n
t
h
term of an AP
715
=
85
+
(
n
−
1
)
5
126
=
n
−
1
⇒
n
=
127
To get the required sum, apply the formula for sum of
n
terms of an AP
S
n
=
n
2
(
a
+
l
)
=
127
2
(
85
+
715
)
=
50800
The value of
1
+
3
+
5
+
7
+
9
+
.
.
.
.
.
.
.
.
.
.
.
.
.
+
25
is:
Report Question
0%
196
0%
625
0%
225
0%
169
Explanation
We take the first term
a
=
1
, last term
l
=
25
, common difference
d
=
2
and the number of terms
=
n
.
Then, since it is an A.P. series,
l
=
a
+
(
n
−
1
)
d
n
=
l
−
a
d
+
1
=
25
−
1
2
+
1
=
25
−
1
+
2
2
=
25
+
1
2
=
26
2
=
13
Let
S
n
be the sum of
n
=
13
terms
∴
S
n
=
n
2
×
(
a
+
l
)
=
13
2
×
(
1
+
25
)
=
13
×
13
=
169.
So the required sum
=
169
.
Ramkali saved
R
s
.
5
in the first week of a year and then increased her weekly savings by
R
s
.
1.75.
If in the nth week, her weekly savings become
R
s
.
20.75
,
find
n
.
Report Question
0%
n
=
10
0%
n
=
12
0%
n
=
16
0%
n
=
20
Explanation
Given that
a
=
5
d
=
1.75
a
n
=
20.75
a
n
=
a
+
(
n
−
1
)
d
20.75
=
5
+
(
n
−
1
)
1.75
15.75
=
(
n
−
1
)
1.75
(
n
−
1
)
=
15.75
1.75
=
1575
175
=
63
7
=
9
∴
n
−
1
=
9
∴
n
=
10
The sum of
3
r
d
and
15
t
h
elements of an arithmetic progression is equal to the sum of
6
t
h
,
11
t
h
a
n
d
13
t
h
elements of the same progression. Then which element of the series should necessarily be equal to zero?
Report Question
0%
1
s
t
0%
9
t
h
0%
12
t
h
0%
None of the above
Explanation
Let the first term of AP be
a
and difference be
d
We know that
n
th term,
a
n
=
a
+
(
n
−
1
)
d
,
where
a
&
d
are the first term amd common difference of an AP.
∴
Then third term will be
=
a
+
2
d
15
t
h
will be
=
a
+
14
d
6
t
h
will be
=
a
+
5
d
11
t
h
will be
=
a
+
10
d
13
t
h
will be
=
a
+
12
d
Then the equation will be
a
+
2
d
+
a
+
14
d
=
a
+
5
d
+
a
+
10
d
+
a
+
12
d
2
a
+
16
d
=
3
a
+
27
d
a
+
11
d
=
0
We understand
a
+
11
d
will be the
12
t
h
term of arithmetic progression.
So, the correct answer is
12
.
The sum of first
n
terms of the
A
.
P
.
√
2
+
√
8
+
√
18
+
√
32
+
.
.
.
.
.
.
.
.
is ......
Report Question
0%
n
(
n
+
1
)
2
0%
2
n
(
n
+
1
)
0%
n
(
n
+
1
)
√
2
0%
1
Explanation
The series can be rewritten as
√
2
+
2
√
2
+
3
√
2
+
4
√
2
+
.
.
.
.
.
So the nth sum
S
n
=
√
2
×
(
1
+
2
+
3
+
.
.
.
.
.
+
n
)
=
√
2
×
n
×
(
n
+
1
)
2
=
n
(
n
+
1
)
√
2
The
9
t
h
term of an AP is
499
and
499
t
h
terms is
9.
The term which is equal to zero is
Report Question
0%
501
t
h
0%
502
t
h
0%
508
t
h
0%
None of these
Explanation
a
9
=
a
+
8
d
a
9
=
499
(
g
i
v
e
n
)
∴
a
+
8
d
=
499.........
(
i
)
a
499
=
a
+
498
d
a
499
=
9
∴
a
+
498
d
=
9.........
(
i
i
)
Subtract (i) from (ii)
⇒
490
d
=
−
490
⇒
d
=
−
1
substitute the value of
d
⇒
a
+
8
(
−
1
)
=
499
⇒
a
=
507
∴
For
a
n
=
0
⇒
a
+
(
n
−
1
)
d
=
0
⇒
507
+
(
n
−
1
)
(
−
1
)
=
0
⇒
507
=
n
−
1
⇒
n
=
508
So,
508
t
h
term is equal to zero.
8
t
h
term of series
2
√
2
+
√
2
+
0
+
.
.
.
.
.
.
will be
Report Question
0%
−
5
√
2
0%
5
√
2
0%
10
√
2
0%
−
10
√
2
Explanation
Given series is :
2
√
2
+
√
2
+
0
+
.
.
.
.
.
.
.
Then,
a
=
2
√
2
,
d
=
(
√
2
−
2
√
2
)
=
−
√
2
a
8
=
a
+
7
d
∴
2
√
2
+
7
×
(
−
√
2
)
∴
a
8
=
−
5
√
2
Find the sum of all whole numbers divisible by
5
but less than
100
.
Report Question
0%
950
0%
925
0%
880
0%
1050
Explanation
All whole numbers divisible by
5
but less then
100
are
0
,
5
,
10
,
15
,
20........95
It is an A.P. with first term, common difference and last term as
0
,
5
and
95
respectively.
a
=
0
,
d
=
5
,
a
n
=
95
a
n
=
a
+
(
n
−
1
)
d
95
=
0
+
(
n
−
1
)
5
5
n
=
100
n
=
20
Hence total numbers are
=
20
Sum of numbers are
S
n
=
n
2
[
a
+
(
n
−
1
)
d
]
S
20
=
20
2
[
0
+
(
20
−
1
)
5
]
S
20
=
10
×
95
=
950
The sum of all natural number divisible by
5
but less than
100
is
950
.
The
4
t
h
term from the end of the AP
−
11
,
−
8
,
−
5
,
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.49
is
Report Question
0%
37
0%
40
0%
43
0%
58
Explanation
Given series of AP:
−
11
,
−
8
,
−
5
,
.
.
.
.
.
.
.49
New series :
49
,
.
.
.
.
.
.
.
.
−
5
,
−
8
,
−
11
In new series of AP first term
a
=
49
,
common difference
d
=
(
−
8
)
−
(
−
5
)
=
−
3
a
4
=
a
+
3
d
=
49
+
3
(
−
3
)
=
49
−
9
=
40
∴
Option B is correct.
What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?
Report Question
0%
530
,
820
0%
179
,
587
0%
560
,
758
0%
164
,
850
Explanation
Step1-Find sequence of an A.P according to question
The lowest 3 digit number is
101
and the highest number is
998
.
Therefore, the AP is
101
,
104
,
107
,
.
.
.
.
.
.
.
,
998
where, the first term is
a
=
101
,
common difference is
d=104 - 101 = 3
and
n
t
h
term is
a
n
=
998
Step2-Find the total number of terms of sequence
Let us find the number of terms is n
We know that the n
t
h
term of AP is given by
a
n
=
a
+
(
n
−
1
)
d
Now, substituting the values, we get:
998
=
101
+
(
n
−
1
)
3
⇒
(
n
−
1
)
3
=
998
−
101
⇒
(
n
−
1
)
3
=
897
⇒
n
−
1
=
897
3
⇒
n
−
1
=
299
⇒
n
=
299
+
1
⇒
n
=
300
Step3-Find sum of 300 terms of an A.P
We know
S
=
n
2
(1st term + last term)
Therefore, we have:
S
=
300
2
(
101
+
998
)
S
=
150
×
1099
S
=
164850
Hence, the sum of all 3 digit numbers that leave a remainder of 2 when divided by 3 is 164,850.
The sum of all odd integers between
2
and
50
divisible by
3
is
Report Question
0%
192
0%
216
0%
168
0%
240
Explanation
The odd intergers between 2 and 50 divisible by 3 are
3
,
9
,
15
,
.
.
.
.
.
.45
This sequence forms an AP with first term
a
=
3
and common difference
d
=
9
−
3
=
6
Since,
l
=
a
+
(
n
−
1
)
d
⇒
45
=
3
+
(
n
−
1
)
6
⇒
42
=
(
n
−
1
)
6
⇒
7
=
n
−
1
⇒
n
=
8
Number of terms in given sequence are 8.
S
n
=
n
2
(
a
+
l
)
⇒
S
n
=
8
2
(
3
+
45
)
⇒
S
n
=
4
(
48
)
⇒
S
n
=
192
∴
Option A is correct.
If the
5
th term of an A.P is eight times the first term and
8
th term exceeds twice the
4
th term by
3
, then the common difference is
Report Question
0%
7
0%
5
0%
6
0%
3
Explanation
n
t
h
term of an AP is given as,
t
n
=
a
+
(
n
−
1
)
d
Where
a
and
d
are first term and common difference respectively.
According to the given condition,
Thus,
t
5
=
a
+
4
d
t
1
=
a
t
8
=
a
+
7
d
t
4
=
a
+
3
d
Given,
t
5
=
8
t
1
⟹
a
+
4
d
=
8
a
⟹
4
d
=
7
a
⟹
d
=
7
a
4
Also given,
t
8
−
2
t
4
=
3
⟹
a
+
7
d
−
2
(
a
+
3
d
)
=
3
⟹
d
−
a
=
3
Substituting
d
=
7
a
4
in the above equation,
7
a
4
−
a
=
3
3
a
4
=
3
we get
a
=
4
and hence,
d
=
7
.
If
9
t
h
term of an A.P. is
47
and
16
t
h
term is
82
, then the general term is
Report Question
0%
9
n
+
2
0%
6
n
−
7
0%
5
n
+
2
0%
none
Explanation
a
9
=
a
+
8
d
=
47
........(1)
a
16
=
a
+
15
d
=
82
.....(2)
Subtract eq (1) by (2)
7
d
=
35
∴
d
=
5
a
+
8
d
=
47
a
+
8
×
5
=
47
∴
a
=
7
Hence general term is
a
n
=
a
+
(
n
−
1
)
d
=
5
n
+
2
Find
a
25
−
a
15
for the A.P
−
6
,
−
10
,
−
14
,
−
18
,
.
.
.
.
.
.
.
.
.
.
Report Question
0%
−
40
0%
−
48
0%
40
0%
−
44
Explanation
Given series is:
−
6
,
−
10
,
−
14
,
−
18.....
a
=
−
6
,
d
=
−
4
As we know
n
t
h
term,
a
n
=
a
+
(
n
−
1
)
d
∴
a
25
=
a
+
24
d
&
a
15
=
a
+
14
d
a
25
−
a
15
=
(
a
+
24
d
)
−
(
a
+
14
d
)
=
10
d
In the following series
d
=
−
4
∴
a
25
−
a
15
=
10
×
(
−
4
)
=
−
40
If an A.P is given by
17
,
14
,
.
.
.
.
.
.
−
40
, then
6
th term from the end is
Report Question
0%
−
22
0%
−
28
0%
−
25
0%
26
Explanation
Given an AP is 17,14...........................-40
Then we get AP is -40,-37,.............14,17
Then d=3,a=-40 n=6
Therefore,
a
6
=
a
+
(
n
−
1
)
d
=
(
−
40
)
+
(
6
−
1
)
3
=
−
40
+
15
=
−
25
,
If three times the
9
th term of an A.P is equal to five times its
13
th term, then which of the following term will be zero.
Report Question
0%
26
th
0%
19
th
0%
21
st
0%
36
th
Explanation
According to the question
3
(
a
+
8
d
)
=
5
(
a
+
12
d
)
⇒
3
a
+
24
d
=
5
a
+
60
d
⇒
2
a
+
36
d
=
0
⇒
a
+
18
d
=
0
⇒
a
+
(
19
−
1
)
d
=
0
Hence
19
t
h
term is zero
The total two-digit numbers which are divisible by
5
, are
Report Question
0%
19
0%
20
0%
17
0%
18
Explanation
10
,
15
,
20
,
.
.
.
.
.
.
.
.
95
n
th term of AP
=
a
+
(
n
−
1
)
d
⇒
95
=
10
+
(
n
−
1
)
5
⇒
95
=
10
+
5
n
−
5
⇒
95
=
5
n
+
5
⇒
n
=
18
Hence, required numbers
=
18
How many terms are there in the sequence
15
1
2
,
13
,
.
.
.
.
.
.
−
47
?
Report Question
0%
27
0%
26
0%
28
0%
None
Explanation
Given series is
15
1
2
,
13
,
.
.
.
.
.
,
−
47
a
n
=
a
+
(
n
−
1
)
d
∴
−
47
=
31
2
+
(
n
−
1
)
(
−
5
2
)
−
47
=
31
2
−
5
n
2
+
5
2
−
47
=
18
−
5
n
2
5
n
2
=
65
∴
n
=
65
×
2
5
=
26
Hence number of terms are
26
Find the
n
t
h
term and, hence, the
25
t
h
term of the following AP:
2
,
5
,
8
,
11
,.........
Report Question
0%
3
n
−
1
,
74
0%
2
n
,
73
0%
3
n
,
75
0%
3
n
+
1
,
72
Explanation
Given series, 2, 5, 8, 11, ... is in AP
Here
First term
a
=
2
Common difference
d
=
5
−
2
=
8
−
5
=
3
We know that, general term
t
n
=
a
+
(
n
−
1
)
d
=
2
+
(
n
−
1
)
3
=
2
+
3
n
−
3
=
3
n
−
1
t
25
=
3
×
25
−
1
=
75
−
1
=
74
If the first, second and last term of an A.P. are
x
,
y
and
2
x
respectively, its sum is
Report Question
0%
x
y
2
(
y
−
x
)
0%
3
x
y
2
(
y
−
x
)
0%
2
x
y
y
−
x
0%
x
y
y
−
x
Explanation
Here
a
1
=
x
,
a
2
=
y
and
a
n
=
2
x
.
∴
First term,
a
=
x
and common difference,
d
=
y
−
x
Since,
a
n
=
a
+
(
n
−
1
)
d
⇒
2
x
=
x
+
(
n
−
1
)
(
y
−
x
)
⇒
x
=
(
n
−
1
)
(
y
−
x
)
⇒
x
y
−
x
=
n
−
1
⇒
x
y
−
x
+
1
=
n
⇒
n
=
y
y
−
x
Now, we know that,
S
n
=
n
2
(
a
+
a
n
)
∴
S
n
=
y
y
−
x
2
(
x
+
2
x
)
⇒
S
n
=
y
y
−
x
2
(
3
x
)
⇒
S
n
=
3
x
y
2
(
y
−
x
)
Hence, option B is correct.
If in an A.P,
a
9
=
0
,
a
19
=
k
, then
a
29
is
Report Question
0%
k
2
0%
10
k
0%
9
k
0%
2
k
Explanation
Let
a
and
d
be the first term and common difference of given AP.
Since,
a
9
=
0
⇒
a
+
8
d
=
0
....(1) [using
a
n
=
a
+
(
n
−
1
)
d
]
Similarly,
a
19
=
k
⇒
a
+
18
d
=
k
....(2)
On subtracting (1) from (2). we get
a
+
18
d
−
a
−
8
d
=
k
⇒
10
d
=
k
⇒
d
=
k
10
By substituting
d
=
k
10
in (1), we get
a
+
8
k
10
=
0
⇒
a
=
−
8
k
10
Now,
a
29
=
a
+
28
d
=
−
8
k
10
+
28
k
10
=
2
k
Hence, option D is correct.
The common difference of
−
2
,
−
4
,
−
6
,
−
8
,
.
.
.
.
.
.
.
.
.
.
.
is :
Report Question
0%
−
2
0%
−
1
0%
2
0%
3
Explanation
Step-1: Write the property of Common difference of an A.P.
Given series: -2,-4,-6,-8...
We know that Common difference = Second term - Fiirst term = Third term - Second term
∴
Common difference=
−
4
−
(
−
2
)
=
−
4
+
2
=
−
2
or - 6 - ( - 4) = - 6 + 4 = - 2
Hence, In the given sequence -2, -4, -6, ......, the common difference is -2, option - A.
The AP whose first term is 10 and common difference is 3 is
Report Question
0%
10, 13, 16, 19, ...
0%
5,7,9, 11, ...
0%
8, 12, 16,20, ...
0%
All the above
Explanation
If
a
1
is the first term, then the nth term is
a
1
+
(
n
−
1
)
d
, where
d
is the common difference
Here
a
1
=
10
,
d
=
3
,
Hence,
a
2
=
a
1
+
(
2
−
1
)
d
=
10
+
3
=
13
a
3
=
a
1
+
(
3
−
1
)
d
=
10
+
6
=
16
and so on.....
Which term of the A.P
5
,
15
,
25
−
−
−
−
−
will be
100
more than its
31
s
t
term?
Report Question
0%
41
s
t
0%
42
n
d
0%
40
t
h
0%
43
r
d
Explanation
The given A.P is
5
,
15
,
25.......
where the first term is
a
=
5
and the common difference
d
will be calculated by subtracting the first term from the second term as shown below:
d
=
15
−
5
=
10
Now,
31
st term
=
a
+
30
d
, substituting the values of
a
and
d
, we get:
31
st term
=
5
+
(
30
×
10
)
=
5
+
300
=
305
It is given that the A.P of the given terms is
100
more than its
31
st term, thus, we have:
a
n
=
305
+
100
=
405
⇒
a
+
(
n
−
1
)
d
=
405
⇒
5
+
(
n
−
1
)
10
=
405
⇒
(
n
−
1
)
10
=
405
−
5
⇒
(
n
−
1
)
10
=
400
⇒
n
−
1
=
400
10
⇒
n
−
1
=
40
⇒
n
=
40
+
1
⇒
n
=
41
Hence, the
41
st term of the given A.P will be
100
more than its
31
st term.
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0
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