# Class10 NCRT Arithmetic Progressions Exercise – 5.1 pdf || UP Board

# __Arithmetic Progressions__

__Exercise – 5.1__

__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 a_{1} = ₹15

Fare for 2 km a_{2} = ₹(15 + 8) = ₹23

Fare for 3 km a_{3} = ₹(23 + 8) = ₹31

Fare for 4 km a_{4} = ₹(31 + 8) = ₹39

a_{2} – a_{1} = 23 –
15 = 8

a_{3} – a_{2} = 31 –
23 = 8

a_{4} – a_{3} = 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 a_{1}
= x

The remaining air in the cylinder
after the first removal a_{2 }= x – ¼ x = ¾ x

The remaining air in the cylinder
after the second removal a_{3} = ¾ x – ¼(¾ x) = 9/16 x

The remaining air in the cylinder
after the third removal a_{4} =

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 a_{1} = ₹150

2 meter digging cost a_{2}
= ₹(150 + 50) = ₹200

3 meter
digging cost a_{3}
= ₹(200 + 50) = ₹250

4 meter
digging cost a_{4} = ₹(250 + 50)
= ₹300

a_{2} – a_{1} = 200
– 150 = 50

a_{3} – a_{2} = 250
– 200 = 50

a_{4} – a_{3} = 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 a_{1} = ₹10000

Amount after 1 year a_{2} = ₹10000(1 + 8/100)^{1}
= ₹ 10800

Amount after 2 years a_{3} = ₹10800(1 +
8/100)^{2} = ₹ 10800(108/100)^{2}

⇒ ₹12597

Amount after 3 years a_{3} = ₹12597(1 +
8/100)^{3} = ₹10800(108/100)^{3}

⇒ ₹15869

a_{2} – a_{1} =
₹10800 – ₹10000 = ₹800

a_{3} – a_{2} =
₹12597 – ₹10800 = ₹1797

a_{4} – a_{3} =
₹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 a_{1} = a = 10

Second term a_{2} = a_{1} + d =
10 + 10 = 20

Third term a_{3} = a_{2} + d =
20 + 10 = 30

Fourth term a_{4} = a_{3} + d =
30 + 10 = 40

(ii) a = -2, d = 0

First term a_{1} = a = -2

Second term a_{2} = a_{1} + d =
-2 + 0 = -2

Third term a_{3} = a_{2} + d =
-2 + 0 = -2

Fourth term a_{4} = a_{3}
+ d = -2 + 0 = -2

(iii) a = 4, d = -3

First term a_{1} = a = 4

Second term a_{2} = a_{1} + d =
4 – 3 = 1

Third term a_{3} = a_{2} + d =
1 – 3 = -2

Fourth term a_{4} = a_{3}
+ d = -2 – 3 = -5

(iv) a = -1, d = ½

First term a_{1} = a = -1

Second term a_{2} = a_{1} + d =
-1 + ½ = - ½

Third term a_{3} = a_{2} + d =
- ½ + ½ = 0

Fourth term a_{4} = a_{3}
+ d = 0 + ½ = ½

(v) a = -1.25, d = -0.25

First term a_{1} = a = -1.25

Second term a_{2} = a_{1} + d =
-1.25 – 0.25 = -1.50

Third term a_{3} = a_{2}
+ d = -1.50 – 0.25 = -1.75

Fourth term a_{4} = a_{3}
+ 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 = a_{2} - a_{1} =
1 – 3 = -2

(ii) -5, -1,
3, 7, ……

First term a = -5

Common difference d = a_{2}
- a_{1} = -1 – (-5) = -1 + 5 = 4

(iii) 1/3,
5/3, 9/3, 13/3,……

First term a = 1/3

Common difference d = a_{2}
- a_{1} = 5/3 – 1/3 = 4/3

(iv) 0.6,
1.7, 2.8, 3.9, ……

First term a = 0.6

Common difference d = a_{2}
- a_{1} = 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,……

a_{2} – a_{1} = 4 –
2 = 2

a_{3} – a_{2} = 8 –
4 = 4

a_{4} – a_{3} = 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,……

a_{2} – a_{1} = 5/2
– 2 = ½

a_{3} – a_{2} = 3 –
5/2 = ½

a_{4} – a_{3} = 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 a_{5} = a_{4} + d =
7/2 + ½ = 4

Sixth term a_{6} = a_{5} + d =
4 + ½ = 9/2

Seventh term a_{7} = a_{6} + d =
9/2 + ½ = 10/2 = 5

(iii) -1.2, -3.2, -5.2, -7.2,…….

a_{2} – a_{1} = -3.2
– (-1.2) = -2.0

a_{3} – a_{2} = -5.2
– (-3.2) = -2.0

a_{4} – a_{3} = -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 a_{5} = a_{4} + d =
-7.2 – 2.0 = -9.2

Sixth term a_{6} = a_{5}
+ d = -9.2 – 2.0 = -11.2

Seventh term a_{7} = a_{6}
+ d = -11.2 – 2.0 = -13.2

(iv) -10, -6, -2, 2,…….

a_{2} – a_{1} = -6 –
(-10) = 4

a_{3} – a_{2} = -2 –
(-6) = 4

a_{4} – a_{3} = 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 a_{5} = a_{4} + d =
2 + 4 = 6

Sixth term a_{6} = a_{5}
+ d = 6 + 4 = 10

Seventh term a_{7} = a_{6}
+ d = 10 + 4 = 14

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …….

a_{2} – a_{1} = (3 +
√2) – 3 = √2

a_{3} – a_{2} = (3 +
2√2) – (3 + √2) = √2

a_{4} – a_{3} = (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 a_{5} = a_{4} + d =
(3 + 3√2) + √2 = 3 + 4√2

Sixth term a_{6} = a_{5}
+ d = (3 + 4√2) + √2 = 3 + 5√2

Seventh term a_{7} = a_{6}
+ d = (3 + 5√2) + √2 = 3 + 6√2

(vi) 0.2, 0.22, 0.222, 0.2222, ………

a_{2} – a_{1} = 0.22
– 0.2 = 0.02

a_{3} – a_{2} =
0.222 – 0.22 = 0.002

a_{4} – a_{3} =
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,…….

a_{2} – a_{1} = -4 –
0 = -4

a_{3} – a_{2} = -8 –
(-4) = -8 + 4 = - 4

a_{4} – a_{3} = -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 a_{5} = a_{4} + d =
-12 - 4 = -16

Sixth term a_{6} = a_{5}
+ d = -16 - 4 = -20

Seventh term a_{7} = a_{6}
+ 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,……

a_{2} – a_{1} = 3 –
1 = 2

a_{3} – a_{2} = 9 –
3 = 6

a_{4} – a_{3} = 27 –
9 = 18

The difference of consecutive terms
is not the same. Hence, it is not an A.P.

(x) a, 2a, 3a, 4a, …….

a_{2} – a_{1} = 2a –
a = a

a_{3} – a_{2} = 3a –
2a = a

a_{4} – a_{3} = 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 a_{5} = a_{4} + d =
4a + a = 5a

Sixth term a_{6} = a_{5} + d =
5a + a = 6a

Seventh term a_{7} = a_{6}
+ d = 6a + a = 7a

(xi) a, a^{2}, a^{3},
a^{4},……..

a_{2} – a_{1} = a^{2}
– a

a_{3} – a_{2} = a^{3}
– a^{2}

a_{4} – a_{3} = a^{4}
– a^{3}

The difference of consecutive terms
is not the same. Hence, it is not an A.P.

(xii) √2, √8, √18, √32,…….

a_{2} – a_{1} = √8 –
√2 = 2√2 - √2 = √2

a_{3} – a_{2} = √18
- √8 = 3√2 - 2√2 = √2

a_{4} – a_{3} = √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 a_{5} = a_{4} + d =
√32 + √2 = 4√2 + √2 = 5√2 = √(25×2) = √50

Sixth term a_{6} = a_{5}
+ d = 5√2 + √2 = 6√2 = √(36×2) = √72

Seventh term a_{7} = a_{6}
+ d = 6√2 + √2 = 7√2 = √(49×2) =√98

(xiii) √3, √6, √9, √12,…….

a_{2} – a_{1} = √6 –
√3

a_{3} – a_{2} = √9 –
√6 = 3 - √6

a_{4} – a_{3} = √12
– √9 = 2√3 - 3

The difference of consecutive terms
is not the same. Hence, it is not an A.P.

(xiv) 1^{2}, 3^{2},
5^{2}, 7^{2},……….

a_{2} – a_{1} = 3^{2}
– 1^{2} = 9 – 1 = 8

a_{3} – a_{2} = 5^{2}
– 3^{2} = 25 – 9 = 16

a_{4} – a_{3} = 7^{2}
– 5^{2} = 49 – 25 = 24

The difference of consecutive terms
is not the same. Hence, it is not an A.P.

(xv) 1^{2}, 5^{2}, 7^{2},
73,………

a_{2} – a_{1} = 5^{2}
– 1^{2} = 25 – 1 = 24

a_{3} – a_{2} = 7^{2}
– 5^{2} = 49 – 25 = 24

a_{4} – a_{3} = 73 –
7^{2} = 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 a_{5} = a_{4} + d =
73 + 24 = 97

Sixth term a_{6} = a_{5}
+ d = 97 + 24 = 121

Seventh term a_{7} = a_{6}
+ d = 121 + 24 = 145

Download Class10 NCRT Arithmetic Progressions Exercise – 5.1 pdf

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