Class10 NCRT Arithmetic Progressions Exercise – 5.1 pdf || UP Board
Arithmetic Progressions
Exercise – 5.1
1. In which of the following
situations, does the list of numbers involved make an arithmetic progression,
and why?
(i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.
Fare for 1 km a1 = ₹15
Fare for 2 km a2 = ₹(15 + 8) = ₹23
Fare for 3 km a3 = ₹(23 + 8) = ₹31
Fare for 4 km a4 = ₹(31 + 8) = ₹39
a2 – a1 = 23 –
15 = 8
a3 – a2 = 31 –
23 = 8
a4 – a3 = 39 –
31 = 8
The difference of consecutive terms
is the same. Hence, it is an A.P.
(ii) The amount of air present in a cylinder when a vacuum pump removes ¼ of the air remaining in the
cylinder at a time.
Let the amount of air present in the
cylinder a1
= x
The remaining air in the cylinder
after the first removal a2 = x – ¼ x = ¾ x
The remaining air in the cylinder
after the second removal a3 = ¾ x – ¼(¾ x) = 9/16 x
The remaining air in the cylinder
after the third removal a4 =
The difference of consecutive terms
is not the same. Hence, it is not an A.P.
(iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50
for each subsequent metre.
1 meter
digging cost a1 = ₹150
2 meter digging cost a2
= ₹(150 + 50) = ₹200
3 meter
digging cost a3
= ₹(200 + 50) = ₹250
4 meter
digging cost a4 = ₹(250 + 50)
= ₹300
a2 – a1 = 200
– 150 = 50
a3 – a2 = 250
– 200 = 50
a4 – a3 = 300
– 250 = 50
The difference of consecutive terms
is the same. Hence, it is an A.P.
(iv) The amount of money in the account every year, when ₹10000 is
deposited at compound interest at 8% per annum.
Rate = 8% per annum, Principal a1 = ₹10000
Amount after 1 year a2 = ₹10000(1 + 8/100)1
= ₹ 10800
Amount after 2 years a3 = ₹10800(1 +
8/100)2 = ₹ 10800(108/100)2
⇒ ₹12597
Amount after 3 years a3 = ₹12597(1 +
8/100)3 = ₹10800(108/100)3
⇒ ₹15869
a2 – a1 =
₹10800 – ₹10000 = ₹800
a3 – a2 =
₹12597 – ₹10800 = ₹1797
a4 – a3 =
₹15869 – ₹12597 = ₹3272
The difference of consecutive terms
is not the same. Hence, it is not an A.P.
2. Write first four terms of the A.P, When the first term a and the common
difference d are given as follows:
(i) a = 10, d = 10
First term a1 = a = 10
Second term a2 = a1 + d =
10 + 10 = 20
Third term a3 = a2 + d =
20 + 10 = 30
Fourth term a4 = a3 + d =
30 + 10 = 40
(ii) a = -2, d = 0
First term a1 = a = -2
Second term a2 = a1 + d =
-2 + 0 = -2
Third term a3 = a2 + d =
-2 + 0 = -2
Fourth term a4 = a3
+ d = -2 + 0 = -2
(iii) a = 4, d = -3
First term a1 = a = 4
Second term a2 = a1 + d =
4 – 3 = 1
Third term a3 = a2 + d =
1 – 3 = -2
Fourth term a4 = a3
+ d = -2 – 3 = -5
(iv) a = -1, d = ½
First term a1 = a = -1
Second term a2 = a1 + d =
-1 + ½ = - ½
Third term a3 = a2 + d =
- ½ + ½ = 0
Fourth term a4 = a3
+ d = 0 + ½ = ½
(v) a = -1.25, d = -0.25
First term a1 = a = -1.25
Second term a2 = a1 + d =
-1.25 – 0.25 = -1.50
Third term a3 = a2
+ d = -1.50 – 0.25 = -1.75
Fourth term a4 = a3
+ d = -1.75 – 0.25 = -2.00
3. For the following A.Ps, write the first term and the common difference:
(i) 3, 1, -1, -3,……
First term a = 3
Common difference d = a2 - a1 =
1 – 3 = -2
(ii) -5, -1,
3, 7, ……
First term a = -5
Common difference d = a2
- a1 = -1 – (-5) = -1 + 5 = 4
(iii) 1/3,
5/3, 9/3, 13/3,……
First term a = 1/3
Common difference d = a2
- a1 = 5/3 – 1/3 = 4/3
(iv) 0.6,
1.7, 2.8, 3.9, ……
First term a = 0.6
Common difference d = a2
- a1 = 1.7 – 0.6 = 1.1
4. Which of the following are APs?
If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16,……
a2 – a1 = 4 –
2 = 2
a3 – a2 = 8 –
4 = 4
a4 – a3 = 16 –
8 = 8
The difference of consecutive terms
is not the same. Hence, it is not an A.P.
(ii) 2, 5/2, 3, 7/2,……
a2 – a1 = 5/2
– 2 = ½
a3 – a2 = 3 –
5/2 = ½
a4 – a3 = 7/2
– 3 = ½
The difference of consecutive terms
is the same. Hence, it is an A.P.
Common difference d = ½, next three
terms of this AP are below-
Fifth term a5 = a4 + d =
7/2 + ½ = 4
Sixth term a6 = a5 + d =
4 + ½ = 9/2
Seventh term a7 = a6 + d =
9/2 + ½ = 10/2 = 5
(iii) -1.2, -3.2, -5.2, -7.2,…….
a2 – a1 = -3.2
– (-1.2) = -2.0
a3 – a2 = -5.2
– (-3.2) = -2.0
a4 – a3 = -7.2
– (-5.2) = -2.0
The difference of consecutive terms
is the same. Hence, it is an A.P.
Common difference d = -2.0, next three
terms of this AP are below-
Fifth term a5 = a4 + d =
-7.2 – 2.0 = -9.2
Sixth term a6 = a5
+ d = -9.2 – 2.0 = -11.2
Seventh term a7 = a6
+ d = -11.2 – 2.0 = -13.2
(iv) -10, -6, -2, 2,…….
a2 – a1 = -6 –
(-10) = 4
a3 – a2 = -2 –
(-6) = 4
a4 – a3 = 2 –
(-2) = 4
The difference of consecutive terms
is the same. Hence, it is an A.P.
Common difference d = 4, next three
terms of this AP are below -
Fifth term a5 = a4 + d =
2 + 4 = 6
Sixth term a6 = a5
+ d = 6 + 4 = 10
Seventh term a7 = a6
+ d = 10 + 4 = 14
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …….
a2 – a1 = (3 +
√2) – 3 = √2
a3 – a2 = (3 +
2√2) – (3 + √2) = √2
a4 – a3 = (3 +
3√2) – (3 + 2√2) = √2
The difference of consecutive terms
is the same. Hence, it is an A.P.
Common difference d = √2, next three
terms of this AP are below -
Fifth term a5 = a4 + d =
(3 + 3√2) + √2 = 3 + 4√2
Sixth term a6 = a5
+ d = (3 + 4√2) + √2 = 3 + 5√2
Seventh term a7 = a6
+ d = (3 + 5√2) + √2 = 3 + 6√2
(vi) 0.2, 0.22, 0.222, 0.2222, ………
a2 – a1 = 0.22
– 0.2 = 0.02
a3 – a2 =
0.222 – 0.22 = 0.002
a4 – a3 =
0.2222 – 0.222 = 0.0002
The difference of consecutive terms
is not the same. Hence, it is not an A.P.
(vii) 0, -4, -8, -12,…….
a2 – a1 = -4 –
0 = -4
a3 – a2 = -8 –
(-4) = -8 + 4 = - 4
a4 – a3 = -12
– (-8) = -12 + 8 = - 4
The difference of consecutive terms
is the same. Hence, it is an A.P.
Common difference d = - 4, next three
terms of this AP are below -
Fifth term a5 = a4 + d =
-12 - 4 = -16
Sixth term a6 = a5
+ d = -16 - 4 = -20
Seventh term a7 = a6
+ d = -20 - 4 = -24
The difference of consecutive terms
is the same. Hence, it is an A.P.
Common difference d = 0, next three
terms of this AP are below -
(ix) 1, 3, 9, 27,……
a2 – a1 = 3 –
1 = 2
a3 – a2 = 9 –
3 = 6
a4 – a3 = 27 –
9 = 18
The difference of consecutive terms
is not the same. Hence, it is not an A.P.
(x) a, 2a, 3a, 4a, …….
a2 – a1 = 2a –
a = a
a3 – a2 = 3a –
2a = a
a4 – a3 = 4a –
3a = a
The difference of consecutive terms
is the same. Hence, it is an A.P.
Common difference d = a, next three
terms of this AP are below -
Fifth term a5 = a4 + d =
4a + a = 5a
Sixth term a6 = a5 + d =
5a + a = 6a
Seventh term a7 = a6
+ d = 6a + a = 7a
(xi) a, a2, a3,
a4,……..
a2 – a1 = a2
– a
a3 – a2 = a3
– a2
a4 – a3 = a4
– a3
The difference of consecutive terms
is not the same. Hence, it is not an A.P.
(xii) √2, √8, √18, √32,…….
a2 – a1 = √8 –
√2 = 2√2 - √2 = √2
a3 – a2 = √18
- √8 = 3√2 - 2√2 = √2
a4 – a3 = √32
- √18 = 4√2 - 3√2 = √2
The difference of consecutive terms
is the same. Hence, it is an A.P.
Common difference d = √2, next three
terms of this AP are below -
Fifth term a5 = a4 + d =
√32 + √2 = 4√2 + √2 = 5√2 = √(25×2) = √50
Sixth term a6 = a5
+ d = 5√2 + √2 = 6√2 = √(36×2) = √72
Seventh term a7 = a6
+ d = 6√2 + √2 = 7√2 = √(49×2) =√98
(xiii) √3, √6, √9, √12,…….
a2 – a1 = √6 –
√3
a3 – a2 = √9 –
√6 = 3 - √6
a4 – a3 = √12
– √9 = 2√3 - 3
The difference of consecutive terms
is not the same. Hence, it is not an A.P.
(xiv) 12, 32,
52, 72,……….
a2 – a1 = 32
– 12 = 9 – 1 = 8
a3 – a2 = 52
– 32 = 25 – 9 = 16
a4 – a3 = 72
– 52 = 49 – 25 = 24
The difference of consecutive terms
is not the same. Hence, it is not an A.P.
(xv) 12, 52, 72,
73,………
a2 – a1 = 52
– 12 = 25 – 1 = 24
a3 – a2 = 72
– 52 = 49 – 25 = 24
a4 – a3 = 73 –
72 = 73 – 49 = 24
The difference of consecutive terms
is the same. Hence, it is an A.P.
Common difference d = 24, next three
terms of this AP are below-
Fifth term a5 = a4 + d =
73 + 24 = 97
Sixth term a6 = a5
+ d = 97 + 24 = 121
Seventh term a7 = a6
+ d = 121 + 24 = 145
Download Class10 NCRT Arithmetic Progressions Exercise – 5.1 pdf
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