RS Aggarwal Solutions: Exercise 5C - Arithmetic Progressions Class 10 Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : RS Aggarwal Solutions: Exercise 5C - Arithmetic Progressions Class 10 Notes | EduRev

 Page 1


1. The first three terms of an AP are respectively (3y 1), (3y + 5) and (5y + 1), find the value
of y.
3 , 3 5 1 y y 5 1 y
Sol: 
The terms and are in AP.
3 3 5 3 y 5
3 5 3 1 5 1 3 5
6 2 4
2 10
5
y y y
y y y y
y
y
y
5 1 1
Hence, the value of y is 5. 
2. 1 , 2 k k 2 1 k
1 , 2 k k 2 1 k
If and are the three successive terms of an AP, find the value of k.
Sol:
It is given that and are the three successive terms of an AP.
2 k 1 k 2 k 1 2 k 1
1 2
3
k
k
Hence, the value of k is 3. 
Page 2


1. The first three terms of an AP are respectively (3y 1), (3y + 5) and (5y + 1), find the value
of y.
3 , 3 5 1 y y 5 1 y
Sol: 
The terms and are in AP.
3 3 5 3 y 5
3 5 3 1 5 1 3 5
6 2 4
2 10
5
y y y
y y y y
y
y
y
5 1 1
Hence, the value of y is 5. 
2. 1 , 2 k k 2 1 k
1 , 2 k k 2 1 k
If and are the three successive terms of an AP, find the value of k.
Sol:
It is given that and are the three successive terms of an AP.
2 k 1 k 2 k 1 2 k 1
1 2
3
k
k
Hence, the value of k is 3. 
 
3. If 18, a, (b - 3) are in AP, then find the value of (2a b)
Sol:
It is given that 18, , 3 a b are in AP.
18 3
18 3
2 15
a b a
a a b
a b
Hence, the required value is 15.
4. If the numbers a, 9, b, 25 from an AP, find a and b.
Sol:
It is given that the numbers ,9, ,25 a b from an AP.
9 9 25 a b b
So,
9 25
2 34
17
b b
b
b
Also,
9 9
18
18 17 17
1
a b
a b
a b
a
Hence, the required values of a and b are 1 and 17, respectively.
5. If the numbers (2n 1), (3n+2) and (6n -1) are in AP, find the value of n and the numbers
Sol:
It is given that the numbers 2 1 , 3 2 n n and 6 1 n are in AP.
3 2 2 1 6 1 3 2
3 2 2 1 6 1 3 2
3 3 3
2 6
3
n n n n
n n n n
n n
n
n
When, 3 n
2 1 2 3 1 6 1 5
3 2 3 3 2 9 2 11
6 1 6 3 1 18 1 17
n
n
n
Hence, the required value of n is 3 and the numbers are 5, 11 and 17.
Page 3


1. The first three terms of an AP are respectively (3y 1), (3y + 5) and (5y + 1), find the value
of y.
3 , 3 5 1 y y 5 1 y
Sol: 
The terms and are in AP.
3 3 5 3 y 5
3 5 3 1 5 1 3 5
6 2 4
2 10
5
y y y
y y y y
y
y
y
5 1 1
Hence, the value of y is 5. 
2. 1 , 2 k k 2 1 k
1 , 2 k k 2 1 k
If and are the three successive terms of an AP, find the value of k.
Sol:
It is given that and are the three successive terms of an AP.
2 k 1 k 2 k 1 2 k 1
1 2
3
k
k
Hence, the value of k is 3. 
 
3. If 18, a, (b - 3) are in AP, then find the value of (2a b)
Sol:
It is given that 18, , 3 a b are in AP.
18 3
18 3
2 15
a b a
a a b
a b
Hence, the required value is 15.
4. If the numbers a, 9, b, 25 from an AP, find a and b.
Sol:
It is given that the numbers ,9, ,25 a b from an AP.
9 9 25 a b b
So,
9 25
2 34
17
b b
b
b
Also,
9 9
18
18 17 17
1
a b
a b
a b
a
Hence, the required values of a and b are 1 and 17, respectively.
5. If the numbers (2n 1), (3n+2) and (6n -1) are in AP, find the value of n and the numbers
Sol:
It is given that the numbers 2 1 , 3 2 n n and 6 1 n are in AP.
3 2 2 1 6 1 3 2
3 2 2 1 6 1 3 2
3 3 3
2 6
3
n n n n
n n n n
n n
n
n
When, 3 n
2 1 2 3 1 6 1 5
3 2 3 3 2 9 2 11
6 1 6 3 1 18 1 17
n
n
n
Hence, the required value of n is 3 and the numbers are 5, 11 and 17.
 
6. How many three-digit natural numbers are divisible by 7?
Sol:
The three digit natural numbers divisible by 7 are 105,112,119......,994
Clearly, these number are in AP.
Here, 105 112 105 7 a a n d d
Let this AP contains n terms. Then,
994
105 1 7 994 1
7 98 994
7 994 98 986
128
n
n
a
n a a n d
n
n
n
Hence, there are 128 three digit numbers divisible by 7.
7. How many three-digit natural numbers are divisible by 9?
Sol:
The three-digit natural numbers divisible by 9 are 108, 11
Clearly, these number are in AP.
Here. a = 108 and d = 117 108 = 9
Let this AP contains n terms. Then,
999
108 1 9 999 1
9 99 999
9 999 99 900
100
n
n
a
n a a n d
n
n
n
Hence, there are 100 three-digit numbers divisible by 9.
8. If the sum of first m terms of an AP is (
2
2 3 m m ) then what is its second term?
Sol:
Let 
m
S denotes the sum of first m terms of the AP.
2
2
2 2
1
2 3
2 1 3 1 2 2 1 3 1 2 1
m
m
S m m
S m m m m m m m
Now,
t h
m term of AP, 
1 m m m
a S S
2 2
2 3 2 1 4 1
m
a m m m m m
Putting 2, m we get
2
4 2 1 9 a
Page 4


1. The first three terms of an AP are respectively (3y 1), (3y + 5) and (5y + 1), find the value
of y.
3 , 3 5 1 y y 5 1 y
Sol: 
The terms and are in AP.
3 3 5 3 y 5
3 5 3 1 5 1 3 5
6 2 4
2 10
5
y y y
y y y y
y
y
y
5 1 1
Hence, the value of y is 5. 
2. 1 , 2 k k 2 1 k
1 , 2 k k 2 1 k
If and are the three successive terms of an AP, find the value of k.
Sol:
It is given that and are the three successive terms of an AP.
2 k 1 k 2 k 1 2 k 1
1 2
3
k
k
Hence, the value of k is 3. 
 
3. If 18, a, (b - 3) are in AP, then find the value of (2a b)
Sol:
It is given that 18, , 3 a b are in AP.
18 3
18 3
2 15
a b a
a a b
a b
Hence, the required value is 15.
4. If the numbers a, 9, b, 25 from an AP, find a and b.
Sol:
It is given that the numbers ,9, ,25 a b from an AP.
9 9 25 a b b
So,
9 25
2 34
17
b b
b
b
Also,
9 9
18
18 17 17
1
a b
a b
a b
a
Hence, the required values of a and b are 1 and 17, respectively.
5. If the numbers (2n 1), (3n+2) and (6n -1) are in AP, find the value of n and the numbers
Sol:
It is given that the numbers 2 1 , 3 2 n n and 6 1 n are in AP.
3 2 2 1 6 1 3 2
3 2 2 1 6 1 3 2
3 3 3
2 6
3
n n n n
n n n n
n n
n
n
When, 3 n
2 1 2 3 1 6 1 5
3 2 3 3 2 9 2 11
6 1 6 3 1 18 1 17
n
n
n
Hence, the required value of n is 3 and the numbers are 5, 11 and 17.
 
6. How many three-digit natural numbers are divisible by 7?
Sol:
The three digit natural numbers divisible by 7 are 105,112,119......,994
Clearly, these number are in AP.
Here, 105 112 105 7 a a n d d
Let this AP contains n terms. Then,
994
105 1 7 994 1
7 98 994
7 994 98 986
128
n
n
a
n a a n d
n
n
n
Hence, there are 128 three digit numbers divisible by 7.
7. How many three-digit natural numbers are divisible by 9?
Sol:
The three-digit natural numbers divisible by 9 are 108, 11
Clearly, these number are in AP.
Here. a = 108 and d = 117 108 = 9
Let this AP contains n terms. Then,
999
108 1 9 999 1
9 99 999
9 999 99 900
100
n
n
a
n a a n d
n
n
n
Hence, there are 100 three-digit numbers divisible by 9.
8. If the sum of first m terms of an AP is (
2
2 3 m m ) then what is its second term?
Sol:
Let 
m
S denotes the sum of first m terms of the AP.
2
2
2 2
1
2 3
2 1 3 1 2 2 1 3 1 2 1
m
m
S m m
S m m m m m m m
Now,
t h
m term of AP, 
1 m m m
a S S
2 2
2 3 2 1 4 1
m
a m m m m m
Putting 2, m we get
2
4 2 1 9 a
 
Hence, the second term of the AP is 9.
9.
Sol:
The given AP is 3 ,5 ,........ a a
Here,
First term, A = a
Common difference, 3 2 D a a a
Sum of the n terms, 
n
S
2
2 1 2 2 1
2 2
2 2 2
2
2
2
n
n n
a n a S A n D
n
a a n a
n
an
an
Hence, the required sum is 
2
. a n
10. What is the 5
th
Sol:
The given AP is 2, 7, 12, ..., 47.
Let us re-write the given AP in reverse order i.e. 47, 42, .., 12, 7, 2.
Now, the 5th term from the end of the given AP is equal to the 5th term from beginning of
the AP 47, 42,.... ,12, 7, 2.
Consider the AP 47, 42,..., 12, 7, 2.
Here, a = 47 and d = 42 47 = 5
5th term of this AP
= 47 + (5 1) ( 5)
= 47 20
= 27
Hence, the 5th term from the end of the given AP is 27.
11. If 
n
a d the value of 
30 20
a a .
Sol:
Here, 2 a and 7 2 5 d
Page 5


1. The first three terms of an AP are respectively (3y 1), (3y + 5) and (5y + 1), find the value
of y.
3 , 3 5 1 y y 5 1 y
Sol: 
The terms and are in AP.
3 3 5 3 y 5
3 5 3 1 5 1 3 5
6 2 4
2 10
5
y y y
y y y y
y
y
y
5 1 1
Hence, the value of y is 5. 
2. 1 , 2 k k 2 1 k
1 , 2 k k 2 1 k
If and are the three successive terms of an AP, find the value of k.
Sol:
It is given that and are the three successive terms of an AP.
2 k 1 k 2 k 1 2 k 1
1 2
3
k
k
Hence, the value of k is 3. 
 
3. If 18, a, (b - 3) are in AP, then find the value of (2a b)
Sol:
It is given that 18, , 3 a b are in AP.
18 3
18 3
2 15
a b a
a a b
a b
Hence, the required value is 15.
4. If the numbers a, 9, b, 25 from an AP, find a and b.
Sol:
It is given that the numbers ,9, ,25 a b from an AP.
9 9 25 a b b
So,
9 25
2 34
17
b b
b
b
Also,
9 9
18
18 17 17
1
a b
a b
a b
a
Hence, the required values of a and b are 1 and 17, respectively.
5. If the numbers (2n 1), (3n+2) and (6n -1) are in AP, find the value of n and the numbers
Sol:
It is given that the numbers 2 1 , 3 2 n n and 6 1 n are in AP.
3 2 2 1 6 1 3 2
3 2 2 1 6 1 3 2
3 3 3
2 6
3
n n n n
n n n n
n n
n
n
When, 3 n
2 1 2 3 1 6 1 5
3 2 3 3 2 9 2 11
6 1 6 3 1 18 1 17
n
n
n
Hence, the required value of n is 3 and the numbers are 5, 11 and 17.
 
6. How many three-digit natural numbers are divisible by 7?
Sol:
The three digit natural numbers divisible by 7 are 105,112,119......,994
Clearly, these number are in AP.
Here, 105 112 105 7 a a n d d
Let this AP contains n terms. Then,
994
105 1 7 994 1
7 98 994
7 994 98 986
128
n
n
a
n a a n d
n
n
n
Hence, there are 128 three digit numbers divisible by 7.
7. How many three-digit natural numbers are divisible by 9?
Sol:
The three-digit natural numbers divisible by 9 are 108, 11
Clearly, these number are in AP.
Here. a = 108 and d = 117 108 = 9
Let this AP contains n terms. Then,
999
108 1 9 999 1
9 99 999
9 999 99 900
100
n
n
a
n a a n d
n
n
n
Hence, there are 100 three-digit numbers divisible by 9.
8. If the sum of first m terms of an AP is (
2
2 3 m m ) then what is its second term?
Sol:
Let 
m
S denotes the sum of first m terms of the AP.
2
2
2 2
1
2 3
2 1 3 1 2 2 1 3 1 2 1
m
m
S m m
S m m m m m m m
Now,
t h
m term of AP, 
1 m m m
a S S
2 2
2 3 2 1 4 1
m
a m m m m m
Putting 2, m we get
2
4 2 1 9 a
 
Hence, the second term of the AP is 9.
9.
Sol:
The given AP is 3 ,5 ,........ a a
Here,
First term, A = a
Common difference, 3 2 D a a a
Sum of the n terms, 
n
S
2
2 1 2 2 1
2 2
2 2 2
2
2
2
n
n n
a n a S A n D
n
a a n a
n
an
an
Hence, the required sum is 
2
. a n
10. What is the 5
th
Sol:
The given AP is 2, 7, 12, ..., 47.
Let us re-write the given AP in reverse order i.e. 47, 42, .., 12, 7, 2.
Now, the 5th term from the end of the given AP is equal to the 5th term from beginning of
the AP 47, 42,.... ,12, 7, 2.
Consider the AP 47, 42,..., 12, 7, 2.
Here, a = 47 and d = 42 47 = 5
5th term of this AP
= 47 + (5 1) ( 5)
= 47 20
= 27
Hence, the 5th term from the end of the given AP is 27.
11. If 
n
a d the value of 
30 20
a a .
Sol:
Here, 2 a and 7 2 5 d
 
30 20
2 30 1 5 2 20 1 5 1
147 97
50
n
a a
a a n d
Hence, the required value is 50.
12. The nth term of an AP is (3n +5 ). Find its common difference.
Sol:
We have
3 5
n
T n
Common difference
2 1
T T
1
2
3 1 5 8
3 2 5 11
T
T
11 8 3 d
Hence, the common difference is 3.
13. The nth term of an AP is (7 4n). Find its common difference.
Sol:
We have
7 4
n
T n
Common difference
2 1
T T
1
2
7 4 1 3
7 4 2 1
1 3 4
T
T
d
Hence, the common difference is . 4
14. Write the next term for the AP 8, 18, 32,.........
Sol:
The given AP is 8, 18, 32,.........
On simplifying the terms, we get:
2 2,3 2,4 2,......
Here, 2 2 a and 3 2 2 2 2 d
Next term, 
4
3 2 2 3 2 5 2 50 T a d
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