**Download NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 – Arithmetic Progressions**

**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 the each additional km.**

**Solution:** Taxi fare for 1^{st} 1 km = ₹15.

Taxi fare for first 2 km = ₹15 + ₹8 = ₹23

Taxi fare for first 3 km = ₹23+₹8 = ₹31

Taxi fare for first 4 km = ₹31+₹8 = ₹39

Therefore, we observe that, 15, 23, 31, 39….. forms an A.P. because the common difference for each km is same i.e. 8.

Hence, the given situation makes an arithmetic progression.

**(ii) The amount of air present in a cylinder when a vacuum pump removes ****of the air remaining in the cylinder at a time.**

**Solution:** Let V be the initial volume of air in the cylinder.

After pump remaining volume of air = V – V = V.

After pumping 2^{nd} time volume of air = V- V = V = V

After pumping 3^{rd} times volume of air = V- V = V

Therefore, we observe that, V, V, V …… not form an arithmetic because common difference is not same.

**(iii) The cost of digging a well after every meter of digging, when it costs ****₹****150 for the first meter and rises by ****₹****50 for each subsequent meter.**

**Solution:**

Cost of digging for 1^{st} meter = ₹150.

Cost of digging for 2^{nd} meter = 150+50 = ₹200

Cost of digging for 3^{rd} meter = 200+50 = ₹250

Cost for 250+50 = ₹300

Therefore, we observe that 150, 200, 250, 300 ……. forms an A.P. because the common difference is same i.e, 50

Hence, the given situation makes an arithmetic progression.

**(iv) The amount of money in account every year, when ****₹****10000 is deposited at compound interest at 8% per annum.**

**Solution:** Here P = 10000, r = 8% t = 1 year

Then amount after 1 year = ^{}= 10000^{}= 10000 x

= 10800

Amount after 2 year is = ^{}= 10000^{}= 10000 x

= 108×108

= 11664

So the interest after 1 year is = 10800-10000 = 800

Interest after 2 year is = 11664-10000 = 1664

Hence the difference is not same for both years. So, the given situation is not arithmetic process.

**2. Write first four term of AP, when the first term a and the common difference d are given as follows:**

**Solution:** We know that if a be the first term and d be the common difference of an AP then AP is given by

a, a+d, a+2d, a+3d, a+4d, ….

**(i) a = 10, d = 10**

**Solution:** The AP is given by

10, 10+10, 10+2×10, 10+3×10, 10+4×10, ….

10, 20, 30, 40, 50, …..

Therefore, the first four terms of this AP are 10, 20, 30, 40.

**(ii) a = -2, d = 0**

**Solution:** The AP is given by

-2, -2+0, -2+2×0, -2+3×0,….

-2, -2, -2, -2,…

Therefore, the first four terms of this AP are -2, -2, -2 and -2

**(iii) a = 4, d = -3**

**Solution:** The AP is given by

4, 4+(-3), 4+2×(-3), 4+3(-3),…

4, 4-3, 4-6, 4-9,…

4, 1, -2, -5,…

Therefore, the first four terms of the AP are 4, 1, -2 and -5.

**(iv) a = -1, d = **

**Solution:** The AP is given by

-1, -1+ , -1+2×, -1+3×,…

-1, , -1+1, -1+ ,…

-1, , 0, , …

Therefore, the first four terms of this AP are -1, , 0 and .

**3. For the following APs write the first term and the common difference:**

**Solution:** We know that if a_{1}, a_{2}, a_{3, }a_{4,}… be an AP then the first term is a_{1} and the common difference is d = (a_{2-} a_{1})

**(i) 3, 1, -1, -3, ….**

**Solution:**First term = 3

Common difference (d) = a

_{2 }– a

_{1}= 1-3 = -2.

**(ii) -5, -1, 3, 7, ….**

**Solution:**First term = -5

Common difference (d) = a

_{2 }– a

_{1}= -1-(-5) = -1+5 = 4.

**(iii) , ****, ****, ****, ….**

**Solution:**

First term =

Common difference (d) = a_{2 }– a_{1} = – =

**(iv) 0.6, 1.7, 2.8, 3.9, ….**

**Solution:**First term = 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 AP, find their common difference d and write three more terms.**

**Solution:** We know that if a_{1}, a_{2}, a_{3, }a_{4,}… be an AP then a_{2}– a_{1 }= a_{3 }– a_{2} = a_{4}– a_{3 }= …= d

**(i) 2, 4, 8, 16,…**

**Solution:**

Common difference d is given by

a_{2}-a_{1} = 4-2 = 2

a_{3}-a_{2 }= 8-4 = 4

a_{2}-a_{1 }≠ a_{3}-a_{2}Hence, the given series does not form an AP.

**(ii) 2, ****, 3, ****, ….**

**Solution: **

Common difference d is given by

d = a_{2}-a_{1}= – 2 =

d = a_{3}-a_{2 }= 3- =

d = a_{4}-a_{3}= – 3 =

a_{2}-a_{1 }= a_{3}-a_{2 }= a_{4}-a_{3} =

Hence, the given series forms an AP.

And three more terms of this AP are given by,

a_{5}= + = 4

a_{6}= 4+ =

a_{7}= + = 5

**(iii) -1.2, -3.2, -5.2, -7.2,…**

**Solution: **Common difference d is given by

d = a_{2}-a_{1}= -3.2-(-1.2) = -3.2+1.2 = -2

d = a_{3}-a_{2 }= -5.2-(-3.2) = -5.2+3.2 = -2

d = a_{4}-a_{3}= -7.2-(-5.2) = -7.2+5.2 = -2

a_{2}-a_{1 }= a_{3}-a_{2 }= a_{4}-a_{3} = -2

Hence, the given series forms an AP.

And three more terms of this AP are given by,

a_{5}= -7.2-2 = -9.2

a_{6}= -9.2-2 = -11.2

a_{7}= -11.2-2 = -13.2

**(iv) -10, -6, -2, 2,….**

**Solution: **Common difference d is given by

a_{2}-a_{1}= -6-(-10) = -6+10 = 4

a_{3}-a_{2 }= -2-(-6) = -2+6 = 4

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

a_{2}-a_{1 }= a_{3}-a_{2 }= a_{4}-a_{3} = 4

Hence, the given series forms an AP.

And three more terms of this AP are given by,

a_{5}= 2+4 = 6

a_{6}= 6+4 = 10

a_{7}= 10+4 = 14

**(v) 3, 3+√2, 3+2√2, 3+3√2,…**

**Solution:** Common difference d is given by

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

a_{2}-a_{1 }= a_{3}-a_{2 }= a_{4}-a_{3} = √2

Hence, the given series forms an AP.

And three more terms of this AP are given by,

a_{5}= 3+3√2+√2 = 3+4√2

a_{6}= 3+4√2+√2 = 3+5√2

a_{7}= 3+5√2+√2 = 3+6√2

**(vi) 0.2, 0.22, 0.222, 0.2222,…**

**Solution:**Common difference d is given by

a

_{2}-a

_{1 }= 0.22-0.2 = 0.02

a

_{3}-a

_{2 }= 0.222-0.22 = 0.002

a

_{2}-a

_{1 }≠ a

_{3}-a

_{2}Hence, the given series does not form an AP.

**(vii) 0, -4, -8, -12, ….**

**Solution:**Common difference d is given by

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

a

_{2}-a

_{1 }= a

_{3}-a

_{2 }= a

_{4}-a

_{3}= -4

Hence, the given series forms an AP.

And three more terms of this AP are given by,

a

_{5}= -12-4 = -16

a

_{6}= -16-4 = -20

a

_{7}= -20-4 = -24

**(viii) ****, ****, ****, ****, …**

**Solution:** Common difference d is given by

a_{2}-a_{1 }= – = 0

a_{3}-a_{2 }= – = 0

a_{2}-a_{1 }= a_{3}-a_{2 }= a_{4}-a_{3} = 0

Hence, the given series forms an AP.

Three more terms of this AP are given by , and .

**(ix) 1, 3, 9, 27,….**

**Solution:**Common difference d is given by

a

_{3}-a

_{2 }= 3-1 = 2

a

_{4}-a

_{3 }= 9-3 = 6

a

_{2}-a

_{1 }≠ a

_{3}-a

_{2}Hence, the given series does not form an AP.

**(x) a, 2a, 3a, 4a, …**

**Solution:**Common difference d is given by

a

_{2}-a

_{1 }= 2a-a = a

a

_{3}-a

_{2 }= 3a-2a = a

a

_{4}-a

_{3 }= 4a-3a = a

a

_{2}-a

_{1 }= a

_{3}-a

_{2 }= a

_{4}-a

_{3}= a

Hence, the given series forms an AP.

And three more terms of this AP are given by,

a

_{5}= 4a+a = 5a

a

_{6}= 5a+a = 6a

a

_{7}= 6a+a = 7a

**(xi) a, a ^{2}, a^{3}, a^{4}, …**

**Solution:** Common difference d is given by

a_{2}-a_{1 }= a^{2}-a

a_{3}-a_{2 }= a^{3}-a^{2}a_{2}-a_{1 }≠ a_{3}-a_{2}Hence, the given series does not form an AP.

**(xii) √2, √8****, √18****, √32****,…**

**Solution:** The given series can be written as √2, 2√2, 3√2, 4√2,…

Common difference d is given by

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

a_{2}-a_{1 }= a_{3}-a_{2 }= a_{4}-a_{3} = √2

Hence, the given series forms an AP.

And three more terms of this AP are given by,

a_{5}= 4√2+√2 = 5√2

a_{6 }= 5√2+√2 = 6√2

a_{7 }= 6√2+√2 = 7√2

**(xiii) √3, √6, √9, √12****,…**

**Solution:**

Common difference d is given by

a_{2}-a_{1 }= √6-√3

a_{3}-a_{2 }= √9-√6

a_{2}-a_{1 }≠ a_{3}-a_{2}Hence, the given series does not form an AP.

**(xiv) 1 ^{2}, 3^{2}, 5^{2}, 7^{2},…**

**Solution:** The given series can be written as 1, 9, 25, 49, …

a_{2}-a_{1 }= 9 – 1 = 8

a_{3}-a_{2 }= 25 – 9 = 16

a_{2}-a_{1 }≠ a_{3}-a_{2}Hence, the given series does not form an AP.

**(xv) 1 ^{2}, 5^{2}, 7^{2}, 73**

**Solution:** The given series can be written as 1, 25, 49, 73, …

Common difference d is given by

a_{2}-a_{1 }= 25-1 =24

a_{3}-a_{2 }= 49-25 = 24

a_{4}-a_{3 }= 73-49 = 24

a_{2}-a_{1 }= a_{3}-a_{2 }= a_{4}-a_{3} = 24

Hence, the given series forms an AP.

And three more terms of this AP are given by,

a_{5}= 73+24 = 97

a_{6 }= 97+24 = 121

a_{7}= 121+24 = 145

**Download NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 – Arithmetic Progressions**

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