SEBA Class 10 Maths Chapter 5 Arithmetic Progressions

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SEBA Class 10 Maths Chapter 5 Arithmetic Progressions

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Arithmetic Progressions

Chapter: 5

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.

Ans: It can be observed that

Taxi fare for 1st 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

Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more

than the preceding term.

(ii)The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

(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. 

Ans: Cost of digging for first metre = 150

Cost of digging for first 2 metres = 150 + 50 = 200

Cost of digging for first 3 metres = 200 + 50 = 250 

Cost of digging for first 4 metres = 250 + 50 = 300 Clearly, 150, 

200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term. 

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

 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

Ans: a = 10, d = 10

Let the series be a₁, a₂, a₃, a₄, a₅

a₁ = a = 10

a₂ = a₁ + d = 10 + 10 = 20

a₃ = a₂ + d = 20 + 10 = 30

a₄ = a₃ + d = 30 + 10 = 40

a₅ = a₄ + d = 40 + 10 = 50 

Therefore, the series will be 10, 20, 30, 40, 50 … First four terms of this A.P. will be 

10, 20, 30, and 40. 

(ii) a = −2, d = 0

Ans: Let the series be  a¹, a², a³, a⁴ …

a₁ = a = −2

a₂= a₁ + d = − 2 + 0 = −2

a₃ = a₂+ d = − 2 + 0 = −2

a₄= a₃ + d = − 2 + 0 = −2

Therefore, the series will be −2, −2, −2, −2 …

First four terms of this A.P. will be −2, −2, −2 and −2. 

(iii) a = 4, d = −3

Ans: Let the series be a₁, a₂, a₃, a₄…

a₁ = a = 4

a₂ = a¹ + d = 4 − 3 = 1

a₃ = a² + d = 1 − 3 = −2

a₄ = a³ + d = − 2 − 3 = −5

Therefore, the series will be 4, 1, −2, −5 …

First four terms of this A.P. will be 4, 1, −2 and −5.

(iv) a = −1, d = 1/2 

(v) a = −1.25, d = −0.25

Ans: Let the series be a₁, a₂^, a₃, a₄ …

a _{1 =} a = −1.25

a₂ = a₁ + d = − 1.25 − 0.25 = −1.50

a₃= a₂+ d = − 1.50 − 0.25 = −1.75

a₄ = a₃ + d = − 1.75 − 0.25 = −2.00

Clearly, the series will be 1.25, −1.50, −1.75, −2.00 …

First four terms of this A.P. will be −1.25, −1.50, −1.75 and −2.00. 

3. For the following A.Ps, write the first term and the common difference.  

(i) 3, 1, −1, −3,… 

Ans: 3, 1, −1, −3 

Here, first term, a = 3

Common difference, d = Second term − First term

= 1 − 3 = −2 

(ii) −5, −1, 3, 7.

Ans: Here, First term, a = 1/3

Common difference, d = 5/3 − 1/3 = 4/3

(iv) 0.6, 1.7, 2.8, 3.9 

Ans: (iv) Here, First term, a = 0.6

Common difference, d = 1.7 − 0.6  = 1.1

4. Which of the following APs? If they form an AP, find the common difference d and write three more terms.

(i) 2, 4, 8, 16…

Ans: Not an AP because t₂ – t₁ = 2 and t₃ – t₂ – 2 = 8 – 4 = 4 

i.e .. , t₂ – t₁ ≠ t₃ – t₂

(ii) 2, 5/2, 3, 7/2

Ans: It is an AP.

a = 2, d = 1/2 

t₅ = 7/2 + 1/2 = 4, t₆ = 4 + 1/2 = 9/2, t₇ = 9/2 + 1/2 = 5

i.e. t₂ – t₁ = t₃ – t₂ = t₄ – t₃ = 1/2 

(iii) -1.2, -3.2, -5.2, -7.2,… 

Ans: It is an AP.

a = -1.2, d = -2 

t₅  = -7.2 – 2 = -9.2, t₆ = -11.2, t₇ = -13.2

(iv) -10, -6, -2, 2,…

Ans: It is an AP.

a = -10, d = 4

t₅ = 6, t₆ = 10, t₇ = 14

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2,…

Ans: It is an AP.

a = 3, d = √2

t₅ = 3 + 3√2 + √2 = 3 + 4√3, t₆ = 3 + 5√3, t₇ = 3 + 6√3

(vi) 0.2, 0.22, 0.222, 0.2222,…

Ans: It is not an AP.

Here, t₂, – t₁ = 0.22 – 0.2 = 0.02…

t₃ – t₂ = 0.222 – 0.22 = 0.002

i.e. t₂ – t₁ ≠ t₃ – t₂ 

(vii) 0, -4, -8, – 12

Ans: It is an AP.

a = 0, d = -4

t₅ = -16, t₆ = -20, t₇ = -24

(ix) 1, 3, 9, 27

Ans: Not an AP

Here, t₂ – t₁ ≠ t₃ – t₂

(x) a, 2a, 3a, 4a

Ans: It is an AP. 

First term = a, common differences = 1a 

t₅ = 5a, t₆ = 6a, t₇ = 7a

(xi) a, a², a³, a⁴,….

Ans: Not an AP if a ≠ 1. 

Here t₂ – t₁ = a² – a = a(1 – a)

t₃ – t₂ = a³ – a² = a²(1 – a)

t₃ – t₂ ≠ t₂ – t₁ when a ≠ 1

It will be an AP if a = 1 

Hence, the given sequence is an AP only when a = 1 

In this case, First term = 1, common difference = 1

(xiv) 1², 3², 7²

(xv) 1², 5², 7², 73…

5. Which of the following list of numbers does not form an A.P.?

(i) 2, 4, 6, 8, 10, …

(ii) 2, 4, 8, 16, …

(iii) 1, −2, −5, −8, …

(iv) 1.8, 2.0, 2.2, 2.4, …

(A) (i)

(B) (ii)

(C) (iii)

(D) (iv) 

Ans: (B) (ii)

6. Consider the statements P to S. They form jumbled links of common difference in A.P. Choose the option that represents the correct increasing sequence of common difference of the A.P.

P : 2, 5/2, 3, 7/2, …

Q : 1.2, 3.2, 5.2, 7.2, …

R : 7, 10 1/2, 14, 17 1/2, …

S : 1/15, 1/12, 1/10, 7/60, …

(A) S, Q, R, P

(B) R, P, Q, S

(C) S, P, Q, R

(D) P, Q, S, R

Ans: (C) S, P, Q, R

7. If p, 2p − 1 and 2p + 1 are three consecutive terms of an A.P., the value of p is

(A) -3

(B) -2

(C) 3

(D) 6

Ans: (C) 3

8. Show that a − b, a, a + b form consecutive terms of an A.P.

Ans: Here, a₁ = a − b, a₂ = a, a₃ = a + b

∴ a₂ − a₁ = a − (a − b)

= a – a + b

= b

And, a₃ − a₂ = (a + b) − a

= b

∴ a₂ − a₁ = a₃ − a₂

I.e. a_k+1 – a_k remains the same.

Therefore, the terms a − b, a and a + b form an arithmetic progression.

Hence, a − b, a, a + b are consecutive terms of an A.P.

Exercise 5.2

1. Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the n^th of the A.P. 

	a	d	n	an
(i)	7	3	8	…..
(ii)	-18	…..	10	0
(iii)	…..	-3	18	-5
(iv)	-18.9	2.5	…..	3.6
(v)	3.5	0	105	…..

Ans: a = 7, d = 3, n = 8, a_n = ?

We know that,

For an A.P. a_n = a + (n − 1) d

= 7 + (8 − 1) 3

= 7 + (7) 3

= 7 + 21 = 28

Hence, a_n = 28

(ii) Given that a = -18, n = 10, a_n = 0, d = ?

We know that,

a_n = a + (n – 1) d

0 = -18 + (10 – 1) d

18 = 9d

D = 18/9 = 2 

Hence, common difference, d = 2 

(iii) Given that d = −3, n = 18, a_n = −5

We know that, a_n = a + (n − 1) d

−5 = a + (18 − 1) (−3)

−5 = a + (17) (−3)

−5 = a − 51

a = 51 − 5 = 46

Hence, a = 46

(iv)  a = −18.9, d = 2.5, a_n = 3.6, n = ?

We know that, a_n= a + (n − 1) d

3.6 = −18.9 + (n − 1) 2.5

3.6 + 18.9 = (n − 1) 2.5

22.5 = (n − 1) 2.5

Hence, n = 10

(v) a = 3.5, d = 0, n = 105, a_n = ?

We know that, a_n = a + (n − 1) d

a_n = 3.5 + (105 − 1) 0

a_n = 3.5 + 104 × 0

a_n = 3.5

Hence, a_n = 3.5 

2. Choose the correct choice in the following and justify:

(i) 30th term of the A.P: 10, 7, 4, …is

(A) 97

(B) 77

(C) -77

(D) -87

Ans: (C) -77

(ii) 11th term of the AP: -3, -1/2,… is

(A) 28

(B) 22

(C) -38

(D) -48 1/2 

Ans: (B) 22

3. In the following APs find the missing term in the boxes.

Ans: (i) For this A.P., a = 2 and a₃ = 26

We know that, a_n = a + (n − 1) d

a₃ = 2 + (3 − 1) d

26 = 2 + 2d

24 = 2d   d = 12

a₂ = 2 + (2 − 1) 12

= 14

Therefore, 14 is the missing term. 

(ii) Let the missing terms in the boxes be x and y.

Then,

x, 13, y and 3 are in A.P.

Since the difference between consecutive terms is constant,

t₂ − t₁ = t₃ − t₂

⇒ 13 − x = y − 13

⇒ x + y = 26      …(i)

Also,

t₃ − t₂ = t₄ − t₃

⇒ y − 13 = 3 − y

⇒ 2y = 16

⇒ y = 16/2

⇒ y = 8

Putting the value of y in equation (i),

We get

⇒ x + 8 = 26

⇒ x = 26 – 8 

∴ x = 18

Therefore, the missing terms in the boxes are 18 and 8 respectively.

(iii) Let the missing terms in the boxes be a₂ and a₃ respectively.

∴ First term, a = 5

Common difference, d = a₂ − 5

= a₃ − a₂

Number of terms, n = 4

4th term, a₄ = 9 1/2  = 19/2

(iv) Let the missing terms in the boxes be a₂, a₃, a₄ and a₅ respectively.

∴ First term, a = −4

Common difference,

d = a₂ − a = a₂a₂ − (−4) = a₂ + 4

Number of terms, n = 6

6th term, a₆ = 6

We know,

a_n = a + (n − 1)d

⇒ a₆ = a + (6 − 1)d

⇒ 6 = −4 + 5(a₂ + 4)

⇒ 6 + 4 = 5(a₂ + 4)

⇒ 10 = 5(a₂ + 4)

⇒ 10/5 = a₂ + 4

⇒  a₂ + 4 = 10/5

⇒ a₂ + 4 = 2

⇒ a₂ = 2 − 4

⇒ a₂ = −2

∴ Common difference,

d = a₂ + 4 = −2 + 4 = 2

∴ Third term, a₃ = a₂ + d = −2 + 2 = 0

Fourth term, a₄ = a₃ + d = 0 + 2 = 2

Fifth term, a₅ = a₄ + d = 2 + 2 = 4

Therefore, the missing terms in the boxes are −2, 0, 2 and 4 respectively. 

(v) Let the missing terms in the boxes be a₁, a₃, a₄ and a₅ respectively.

∴ First term, a = a₁

Second term, a₂ = 38

Common difference,

d = a₂ − a₁ = 38 − a₁

Number of terms, n = 6

6th term, a₆ = −22

We know,

a_n = a + (n − 1)d

⇒ a₆ = a₁ + (6 − 1)(38 − a₁)

⇒ −22 = a₁ + 5(38 − a₁)

⇒ −22 = a₁ + 190 − 5a₁

⇒ −22 − 190 = −4a₁a₁

⇒ −212 = −4a₁

⇒ 4a₁ = 212

⇒ a₁ = 212/4

⇒ a₁ = 53

∴ Common difference, d = 38 − a₁  = 38 − 53  = −15

∴ Third term,

a₃ = 38 + (−15) = 23

Fourth term, a₄ = 23 + (−15) = 8

Fifth term, a₅ = 8 + (−15)

= −7

Therefore, the missing terms are 53, 23, 8 and −7 respectively.

4. Which term of the A.P. 3, 8, 13, 18, … is 78?

Ans: 3, 8, 13, 18, For this A.P., a = 3 and d = a₂ − a₁ = 8 − 3 = 5

Let n^th  term of this A.P. be 78.

a_n = a + (n − 1)d

78 = 3 + (n − 1) 5

75 = (n − 1) 5

(n − 1) = 15

n = 16

Hence, 16^th term of this A.P. is 78.

5. Find the number of terms in each of the following A.P: 

(i) 7, 13, 19, …, 205 

Ans: 7, 13, 19, …, 205

For this A.P., a = 7 and d = a₂ − a₁= 13 − 7 = 6

Let there are n terms in this A.P. a_n = 205

We know that an = a + (n − 1)d

Therefore, 205 = 7 + (n − 1) 6

198 = (n − 1) 6

33 = (n − 1)

n = 34

Therefore, this given series has 34 terms in it.

(ii) 18, 15 1/2, …, -47

Ans: a = 18, d = -5/2

6. Check whether -150 is a term of the A.P. 11, 8, 5, 2, … 

Ans: a = 11, d = -3

Let if possible t_n = -150

7. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.

Ans: Given that, a₁₁ = 38 and a₁₆ = 73

We know that, a_n = a + (n − 1)d

a₁₁ = a + (11 − 1)d

38 = a + 10d …(1)

Similarly, a₁₆ = a + (16 − 1)d

73 = a + 15d …(2)

On subtracting (1) from (2), we obtain

35 = 5d

d = 7

From equation (1),

38 = a + 10 × (7) 

38 − 70 = a

a = −32

a₃₁ = a + (31 − 1)d

= −32 + 30 (7)

= −32 + 210

= 178

Hence, 31st term is 178. 

8. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term 

Ans: Given that, a₃ = 12 and a₅₀ = 106

We know that, a_n = a + (n − 1)d

a₃ = a + (3 − 1)d

12 = a + 2d …………………………..(I)

Similarly, a₅₀ = a + (50 − 1) d

106 = a + 49d ………………………(II)

On subtracting (I) from (II), we obtain

94 = 47d

d = 2

From equation (I), we obtain

12 = a + 2 (2)

a = 12 − 4 = 8

a₂₉ = a + (29 − 1) d

a₂₉ = 8 + (28)²

a₂₉ = 8 + 56 = 64

Therefore, 29^th term is 64.