NCERT Solutions for Class 10 Math Chapter 5 - Arithmetic Progressions

Answer:

(i) It can be observed that

Taxi fare for 1^st 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) Let the initial volume of air in a cylinder be V lit. In each stroke, the vacuum pump removes of air remaining in the cylinder at a time. In other words, after every stroke, only part of air will remain.

Therefore, volumes will be  

Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

(iii) 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) We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be after n years.

Therefore, after every year, our money will be

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

Answer:

(i) 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

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

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 =

Let the series be a₁, a₂, a₃, a₄ …

Clearly, the series will be

………….

First four terms of this A.P. will be .

(v) a = −1.25, d = −0.25

Let the series be a₁, a₂, a₃, a₄ …

a₁ = 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.

Answer:

(i) 3, 1, −1, −3 …

Here, first term, a = 3

Common difference, d = Second term − First term

= 1 − 3 = −2

(ii) −5, −1, 3, 7 …

Here, first term, a = −5

Common difference, d = Second term − First term

= (−1) − (−5) = − 1 + 5 = 4

(iii)

Here, first term,

Common difference, d = Second term − First term

(iv) 0.6, 1.7, 2.8, 3.9 …

Here, first term, a = 0.6

Common difference, d = Second term − First term

= 1.7 − 0.6

= 1.1

Answer:

(i) 2, 4, 8, 16 …

It can be observed that

a₂ − a₁ = 4 − 2 = 2

a₃ − a₂ = 8 − 4 = 4

a₄ − a₃ = 16 − 8 = 8

i.e., a_k+1− a_k is not the same every time. Therefore, the given numbers are not forming an A.P.

(ii)

It can be observed that

i.e., a_k+1− a_k is same every time.

Therefore, and the given numbers are in A.P.

Three more terms are

(iii) −1.2, −3.2, −5.2, −7.2 …

It can be observed that

a₂ − a₁ = (−3.2) − (−1.2) = −2

a₃ − a₂ = (−5.2) − (−3.2) = −2

a₄ − a₃ = (−7.2) − (−5.2) = −2

i.e., a_k+1− a_k is same every time. Therefore, d = −2

The given numbers are in A.P.

Three more terms are

a₅ = − 7.2 − 2 = −9.2

a₆ = − 9.2 − 2 = −11.2

a₇ = − 11.2 − 2 = −13.2

(iv) −10, −6, −2, 2 …

It can be observed that

a₂ − a₁ = (−6) − (−10) = 4

a₃ − a₂ = (−2) − (−6) = 4

a₄ − a₃ = (2) − (−2) = 4

i.e., a_k+1 − a_k is same every time. Therefore, d = 4

The given numbers are in A.P.

Three more terms are

a₅ = 2 + 4 = 6

a₆ = 6 + 4 = 10

a₇ = 10 + 4 = 14

(v)

It can be observed that

i.e., a_{k+1 −} a_k is same every time. Therefore,

The given numbers are in A.P.

Three more terms are

(vi) 0.2, 0.22, 0.222, 0.2222 ….

It can be observed that

a₂ − a₁ = 0.22 − 0.2 = 0.02

a₃ − a₂ = 0.222 − 0.22 = 0.002

a₄ − a₃ = 0.2222 − 0.222 = 0.0002

i.e., a_k+1 − a_k is not the same every time.

Therefore, the given numbers are not in A.P.

(vii) 0, −4, −8, −12 …

It can be observed that

a₂ − a₁ = (−4) − 0 = −4

a₃ − a₂ = (−8) − (−4) = −4

a₄ − a₃ = (−12) − (−8) = −4

i.e., a_k+1 − a_k is same every time. Therefore, d = −4

The given numbers are in A.P.

Three more terms are

a₅ = − 12 − 4 = −16

a₆ = − 16 − 4 = −20

a₇ = − 20 − 4 = −24

(viii)

It can be observed that

i.e., a_k+1 − a_k is same every time. Therefore, d = 0

The given numbers are in A.P.

Three more terms are

(ix) 1, 3, 9, 27 …

It can be observed that

a₂ − a₁ = 3 − 1 = 2

a₃ − a₂ = 9 − 3 = 6

a₄ − a₃ = 27 − 9 = 18

i.e., a_k+1 − a_k is not the same every time.

Therefore, the given numbers are not in A.P.

(x) a, 2a, 3a, 4a …

It can be observed that

a₂ − a₁ = 2a − a = a

a₃ − a₂ = 3a − 2a = a

a₄ − a₃ = 4a − 3a = a

i.e., a_k+1 − a_k is same every time. Therefore, d = a

The given numbers are in A.P.

Three more terms are

a₅ = 4a + a = 5a

a₆ = 5a + a = 6a

a₇ = 6a + a = 7a

(xi) a, a², a³, a⁴ …

It can be observed that

a₂ − a₁ = a² − a = a (a − 1)

a₃ − a₂ = a³ − a² = a² (a − 1)

a₄ − a₃ = a⁴ − a³ = a³ (a − 1)

i.e., a_k+1 − a_k is not the same every time.

Therefore, the given numbers are not in A.P.

(xii)

It can be observed that

i.e., a_k+1 − a_k is same every time.

Therefore, the given numbers are in A.P.

And,

Three more terms are

(xiii)

It can be observed that

i.e., a_k+1 − a_k is not the same every time.

Therefore, the given numbers are not in A.P.

(xiv) 1², 3², 5², 7² …

Or, 1, 9, 25, 49 …..

It can be observed that

a₂ − a₁ = 9 − 1 = 8

a₃ − a₂ = 25 − 9 = 16

a₄ − a₃ = 49 − 25 = 24

i.e., a_k+1 − a_k is not the same every time.

Therefore, the given numbers are not in A.P.

(xv) 1², 5², 7², 73 …

Or 1, 25, 49, 73 …

It can be observed that

a₂ − a₁ = 25 − 1 = 24

a₃ − a₂ = 49 − 25 = 24

a₄ − a₃ = 73 − 49 = 24

i.e., a_k+1 − a_k is same every time.

Therefore, the given numbers are in A.P.

And, d = 24

Three more terms are

a₅ = 73+ 24 = 97

a₆ = 97 + 24 = 121

a₇ = 121 + 24 = 145

Answer:

I. 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

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

Answer:

I. Given that

A.P. 10, 7, 4, …

First term, a = 10

Common difference, d = a₂ − a₁ = 7 − 10

= −3

We know that, a_n = a + (n − 1) d

a₃₀ = 10 + (30 − 1) (−3)

a₃₀ = 10 + (29) (−3)

a₃₀ = 10 − 87 = −77

Hence, the correct answer is C.

II. Given that, A.P.

First term a = −3

Common difference, d = a₂ − a₁

We know that,

Hence, the answer is B.

Answer:

I.

For this A.P.,

a = 2

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.

For this A.P.,

a₂ = 13 and

a₄ = 3

We know that, a_n = a + (n − 1) d

a₂ = a + (2 − 1) d

13 = a + d (I)

a₄ = a + (4 − 1) d

3 = a + 3d (II)

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

−10 = 2d

d = −5

From equation (I), we obtain

13 = a + (−5)

a = 18

a₃ = 18 + (3 − 1) (−5)

= 18 + 2 (−5) = 18 − 10 = 8

Therefore, the missing terms are 18 and 8 respectively.

III.

For this A.P.,

We know that,

Therefore, the missing terms are and 8 respectively.

IV.

For this A.P.,

a = −4 and

a₆ = 6

We know that,

a_n = a + (n − 1) d

a₆ = a + (6 − 1) d

6 = − 4 + 5d

10 = 5d

d = 2

a₂ = a + d = − 4 + 2 = −2

a₃ = a + 2d = − 4 + 2 (2) = 0

a₄ = a + 3d = − 4 + 3 (2) = 2

a₅ = a + 4d = − 4 + 4 (2) = 4

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

V.

For this A.P.,

a₂ = 38

a₆ = −22

We know that

a_n = a + (n − 1) d

a₂ = a + (2 − 1) d

38 = a + d (1)

a₆ = a + (6 − 1) d

−22 = a + 5d (2)

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

− 22 − 38 = 4d

−60 = 4d

d = −15

a = a₂ − d = 38 − (−15) = 53

a₃ = a + 2d = 53 + 2 (−15) = 23

a₄ = a + 3d = 53 + 3 (−15) = 8

a₅ = a + 4d = 53 + 4 (−15) = −7

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

Answer:

3, 8, 13, 18, …

For this A.P.,

a = 3

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.

Answer:

I. 7, 13, 19, …, 205

For this A.P.,

a = 7

d = a₂ − a₁ = 13 − 7 = 6

Let there are n terms in this A.P.

a_n = 205

We know that

a_n = 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.

For this A.P.,

Let there are n terms in this A.P.

Therefore, a_n = −47 and we know that,

Therefore, this given A.P. has 27 terms in it.

Answer:

For this A.P.,

a = 11

d = a₂ − a₁ = 8 − 11 = −3

Let −150 be the n^th term of this A.P.

We know that,

Clearly, n is not an integer.

Therefore, −150 is not a term of this A.P.

Answer:

Given that,

a₁₁ = 38

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, 31^st term is 178.

Answer:

Given that,

a₃ = 12

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.

Answer:

Given that,

a₃ = 4

a₉ = −8

We know that,

a_n = a + (n − 1) d

a₃ = a + (3 − 1) d

4 = a + 2d (I)

a₉ = a + (9 − 1) d

−8 = a + 8d (II)

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

−12 = 6d

d = −2

From equation (I), we obtain

4 = a + 2 (−2)

4 = a − 4

a = 8

Let n^th term of this A.P. be zero.

a_n = a + (n − 1) d

0 = 8 + (n − 1) (−2)

0 = 8 − 2n + 2

2n = 10

n = 5

Hence, 5^th term of this A.P. is 0.

Answer:

We know that,

For an A.P., a_n = a + (n − 1) d

a₁₇ = a + (17 − 1) d

a₁₇ = a + 16d

Similarly, a₁₀ = a + 9d

It is given that

a₁₇ − a₁₀ = 7

(a + 16d) − (a + 9d) = 7

7d = 7

d = 1

Therefore, the common difference is 1.

Answer:

Given A.P. is 3, 15, 27, 39, …

a = 3

d = a₂ − a₁ = 15 − 3 = 12

a₅₄ = a + (54 − 1) d

= 3 + (53) (12)

= 3 + 636 = 639

132 + 639 = 771

We have to find the term of this A.P. which is 771.

Let n^th term be 771.

a_n = a + (n − 1) d

771 = 3 + (n − 1) 12

768 = (n − 1) 12

(n − 1) = 64

n = 65

Therefore, 65^th term was 132 more than 54^th term.

Alternatively,

Let n^th term be 132 more than 54^th term.

Answer:

Let the first term of these A.P.s be a₁ and a₂ respectively and the common difference of these A.P.s be d.

For first A.P.,

a₁₀₀ = a₁ + (100 − 1) d

= a₁ + 99d

a₁₀₀₀ = a₁ + (1000 − 1) d

a₁₀₀₀ = a₁ + 999d

For second A.P.,

a₁₀₀ = a₂ + (100 − 1) d

= a₂ + 99d

a₁₀₀₀ = a₂ + (1000 − 1) d

= a₂ + 999d

Given that, difference between

100^th term of these A.P.s = 100

Therefore, (a₁ + 99d) − (a₂ + 99d) = 100

a₁ − a₂ = 100 (1)

Difference between 1000^th terms of these A.P.s

(a₁ + 999d) − (a₂ + 999d) = a₁ − a₂

From equation (1),

This difference, a₁ − a₂ = 100

Hence, the difference between 1000^th terms of these A.P. will be 100.

Answer:

First three-digit number that is divisible by 7 = 105

Next number = 105 + 7 = 112

Therefore, 105, 112, 119, …

All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.

The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.

The series is as follows.

105, 112, 119, …, 994

Let 994 be the nth term of this A.P.

a = 105

d = 7

a_n = 994

n = ?

a_n = a + (n − 1) d

994 = 105 + (n − 1) 7

889 = (n − 1) 7

(n − 1) = 127

n = 128

Therefore, 128 three-digit numbers are divisible by 7.

Answer:

First multiple of 4 that is greater than 10 is 12. Next will be 16.

Therefore, 12, 16, 20, 24, …

All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.

When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.

The series is as follows.

12, 16, 20, 24, …, 248

Let 248 be the n^th term of this A.P.

Therefore, there are 60 multiples of 4 between 10 and 250.

Answer:

63, 65, 67, …

a = 63

d = a₂ − a₁ = 65 − 63 = 2

n^th term of this A.P. = a_n = a + (n − 1) d

a_n= 63 + (n − 1) 2 = 63 + 2n − 2

a_n = 61 + 2n (1)

3, 10, 17, …

a = 3

d = a₂ − a₁ = 10 − 3 = 7

n^th term of this A.P. = 3 + (n − 1) 7

a_n = 3 + 7n − 7

a_n = 7n − 4 (2)

It is given that, n^th term of these A.P.s are equal to each other.

Equating both these equations, we obtain

61 + 2n = 7n − 4

61 + 4 = 5n

5n = 65

n = 13

Therefore, 13^th terms of both these A.P.s are equal to each other.

Answer:

=a₃ = 16

a + (3 − 1) d = 16

a + 2d = 16 (1)

a₇ − a₅ = 12

[a+ (7 − 1) d] − [a + (5 − 1) d]= 12

(a + 6d) − (a + 4d) = 12

2d = 12

d = 6

From equation (1), we obtain

a + 2 (6) = 16

a + 12 = 16

a = 4

Therefore, A.P. will be

4, 10, 16, 22, …

Answer:

Given A.P. is

3, 8, 13, …, 253

Common difference for this A.P. is 5.

Therefore, this A.P. can be written in reverse order as

253, 248, 243, …, 13, 8, 3

For this A.P.,

a = 253

d = 248 − 253 = −5

n = 20

a₂₀ = a + (20 − 1) d

a₂₀ = 253 + (19) (−5)

a₂₀ = 253 − 95

a = 158

Therefore, 20^th term from the last term is 158.

Answer:

We know that,

a_n = a + (n − 1) d

a₄ = a + (4 − 1) d

a₄ = a + 3d

Similarly,

a₈ = a + 7d

a₆ = a + 5d

a₁₀ = a + 9d

Given that, a₄ + a₈ = 24

a + 3d + a + 7d = 24

2a + 10d = 24

a + 5d = 12 (1)

a₆ + a₁₀ = 44

a + 5d + a + 9d = 44

2a + 14d = 44

a + 7d = 22 (2)

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

2d = 22 − 12

2d = 10

d = 5

From equation (1), we obtain

a + 5d = 12

a + 5 (5) = 12

a + 25 = 12

a = −13

a₂ = a + d = − 13 + 5 = −8

a₃ = a₂ + d = − 8 + 5 = −3

Therefore, the first three terms of this A.P. are −13, −8, and −3.

Answer:

It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.

Therefore, the salaries of each year after 1995 are

5000, 5200, 5400, …

Here, a = 5000

d = 200

Let after n^th year, his salary be Rs 7000.

Therefore, a_n = a + (n − 1) d

7000 = 5000 + (n − 1) 200

200(n − 1) = 2000

(n − 1) = 10

n = 11

Therefore, in 11th year, his salary will be Rs 7000.

Answer:

Given that,

a = 5

d = 1.75

a_n = 20.75

n = ?

a_n = a + (n − 1) d

n − 1 = 9

n = 10

Hence, n is 10.

Answer:

(i)2, 7, 12 ,…, to 10 terms

For this A.P.,

a = 2

d = a₂ − a₁ = 7 − 2 = 5

n = 10

We know that,

(ii)−37, −33, −29 ,…, to 12 terms

For this A.P.,

a = −37

d = a₂ − a₁ = (−33) − (−37)

= − 33 + 37 = 4

n = 12

We know that,

(iii) 0.6, 1.7, 2.8 ,…, to 100 terms

For this A.P.,

a = 0.6

d = a₂ − a₁ = 1.7 − 0.6 = 1.1

n = 100

We know that,

(iv). …….. , to 11 terms

For this A.P.,

n = 11

We know that,

Answer:

(i)7 + + 14 + …………+ 84

For this A.P.,

a = 7

l = 84

Let 84 be the n^th term of this A.P.

l = a + (n − 1)d

22 = n − 1

n = 23

We know that,

(ii)34 + 32 + 30 + ……….. + 10

For this A.P.,

a = 34

d = a₂ − a₁ = 32 − 34 = −2

l = 10

Let 10 be the n^th term of this A.P.

l = a + (n − 1) d

10 = 34 + (n − 1) (−2)

−24 = (n − 1) (−2)

12 = n − 1

n = 13

(iii)(−5) + (−8) + (−11) + ………… + (−230)

For this A.P.,

a = −5

l = −230

d = a₂ − a₁ = (−8) − (−5)

= − 8 + 5 = −3

Let −230 be the n^th term of this A.P.

l = a + (n − 1)d

−230 = − 5 + (n − 1) (−3)

−225 = (n − 1) (−3)

(n − 1) = 75

n = 76

And,

Answer:

(i) Given that, a = 5, d = 3, a_n = 50

As a_n = a + (n − 1)d,

∴ 50 = 5 + (n − 1)3

45 = (n − 1)3

15 = n − 1

n = 16

(ii) Given that, a = 7, a₁₃ = 35

As a_n = a + (n − 1) d,

∴ a₁₃ = a + (13 − 1) d

35 = 7 + 12 d

35 − 7 = 12d

28 = 12d

(iii)Given that, a₁₂ = 37, d = 3

As a_n = a + (n − 1)d,

a₁₂ = a + (12 − 1)3

37 = a + 33

a = 4

(iv) Given that, a₃ = 15, S₁₀ = 125

As a_n = a + (n − 1)d,

a₃ = a + (3 − 1)d

15 = a + 2d (i)

On multiplying equation (1) by 2, we obtain

30 = 2a + 4d (iii)

On subtracting equation (iii) from (ii), we obtain

−5 = 5d

d = −1

From equation (i),

15 = a + 2(−1)

15 = a − 2

a = 17

a₁₀ = a + (10 − 1)d

a₁₀ = 17 + (9) (−1)

a₁₀ = 17 − 9 = 8

(v)Given that, d = 5, S₉ = 75

As ,

25 = 3(a + 20)

25 = 3a + 60

3a = 25 − 60

a_n = a + (n − 1)d

a₉ = a + (9 − 1) (5)

(vi) Given that, a = 2, d = 8, S_n = 90

As,

90 = n [2 + (n − 1)4]

90 = n [2 + 4n − 4]

90 = n (4n − 2) = 4n² − 2n

4n² − 2n − 90 = 0

4n² − 20n + 18n − 90 = 0

4n (n − 5) + 18 (n − 5) = 0

(n − 5) (4n + 18) = 0

Either n − 5 = 0 or 4n + 18 = 0

n = 5 or

However, n can neither be negative nor fractional.

Therefore, n = 5

a_n = a + (n − 1)d

a₅ = 2 + (5 − 1)8

= 2 + (4) (8)

= 2 + 32 = 34

(vii) Given that, a = 8, a_n = 62, S_n = 210

n = 6

a_n = a + (n − 1)d

62 = 8 + (6 − 1)d

62 − 8 = 5d

54 = 5d

(viii) Given that, a_n = 4, d = 2, S_n = −14

a_n = a + (n − 1)d

4 = a + (n − 1)2

4 = a + 2n − 2

a + 2n = 6

a = 6 − 2n (i)

−28 = n (a + 4)

−28 = n (6 − 2n + 4) {From equation (i)}

−28 = n (− 2n + 10)

−28 = − 2n² + 10n

2n² − 10n − 28 = 0

n² − 5n −14 = 0

n² − 7n + 2n − 14 = 0

n (n − 7) + 2(n − 7) = 0

(n − 7) (n + 2) = 0

Either n − 7 = 0 or n + 2 = 0

n = 7 or n = −2

However, n can neither be negative nor fractional.

Therefore, n = 7

From equation (i), we obtain

a = 6 − 2n

a = 6 − 2(7)

= 6 − 14

= −8

(ix)Given that, a = 3, n = 8, S = 192

192 = 4 [6 + 7d]

48 = 6 + 7d

42 = 7d

d = 6

(x)Given that, l = 28, S = 144 and there are total of 9 terms.

(16) × (2) = a + 28

32 = a + 28

a = 4

Answer:

Let there be n terms of this A.P.

For this A.P., a = 9

d = a₂ − a₁ = 17 − 9 = 8

636 = n [9 + 4n − 4]

636 = n (4n + 5)

4n² + 5n − 636 = 0

4n² + 53n − 48n − 636 = 0

n (4n + 53) − 12 (4n + 53) = 0

(4n + 53) (n − 12) = 0

Either 4n + 53 = 0 or n − 12 = 0

or n = 12

n cannot be . As the number of terms can neither be negative nor fractional, therefore, n = 12 only.

Answer:

Given that,

a = 5

l = 45

S_n = 400

n = 16

l = a + (n − 1) d

45 = 5 + (16 − 1) d

40 = 15d

Answer:

Given that,

a = 17

l = 350

d = 9

Let there be n terms in the A.P.

l = a + (n − 1) d

350 = 17 + (n − 1)9

333 = (n − 1)9

(n − 1) = 37

n = 38

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

Answer:

d = 7

a₂₂ = 149

S₂₂ = ?

a_n = a + (n − 1)d

a₂₂ = a + (22 − 1)d

149 = a + 21 × 7

149 = a + 147

a = 2

Answer:

Given that,

a₂ = 14

a₃ = 18

d = a₃ − a₂ = 18 − 14 = 4

a₂ = a + d

14 = a + 4

a = 10

= 5610

Answer:

Given that,

S₇ = 49

S₁₇ = 289

7 = (a + 3d)

a + 3d = 7 (i)

Similarly,

17 = (a + 8d)

a + 8d = 17 (ii)

Subtracting equation (i) from equation (ii),

5d = 10

d = 2

From equation (i),

a + 3(2) = 7

a + 6 = 7

a = 1

= n²

 

Answer:

(i) a_n = 3 + 4n

a₁ = 3 + 4(1) = 7

a₂ = 3 + 4(2) = 3 + 8 = 11

a₃ = 3 + 4(3) = 3 + 12 = 15

a₄ = 3 + 4(4) = 3 + 16 = 19

It can be observed that

a₂ − a₁ = 11 − 7 = 4

a₃ − a₂ = 15 − 11 = 4

a₄ − a₃ = 19 − 15 = 4

i.e., a_{k + 1} − a_k is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.

= 15 × 35

= 525

(ii) a_n = 9 − 5n

a₁ = 9 − 5 × 1 = 9 − 5 = 4

a₂ = 9 − 5 × 2 = 9 − 10 = −1

a₃ = 9 − 5 × 3 = 9 − 15 = −6

a₄ = 9 − 5 × 4 = 9 − 20 = −11

It can be observed that

a₂ − a₁ = − 1 − 4 = −5

a₃ − a₂ = − 6 − (−1) = −5

a₄ − a₃ = − 11 − (−6) = −5

i.e., a_{k + 1} − a_k is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.

= −465

 

Answer:

Given that,

S_n = 4n − n²

First term, a = S₁ = 4(1) − (1)² = 4 − 1 = 3

Sum of first two terms = S₂

= 4(2) − (2)² = 8 − 4 = 4

Second term, a₂ = S₂ − S₁ = 4 − 3 = 1

d = a₂ − a = 1 − 3 = −2

a_n = a + (n − 1)d

= 3 + (n − 1) (−2)

= 3 − 2n + 2

= 5 − 2n

Therefore, a₃ = 5 − 2(3) = 5 − 6 = −1

a₁₀ = 5 − 2(10) = 5 − 20 = −15

Hence, the sum of first two terms is 4. The second term is 1. 3^rd, 10^th, and n^th terms are −1, −15, and 5 − 2n respectively.

Answer:

The positive integers that are divisible by 6 are

6, 12, 18, 24 …

It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.

a = 6

d = 6

S₄₀ =?

= 20[12 + (39) (6)]

= 20(12 + 234)

= 20 × 246

= 4920

Answer:

The multiples of 8 are

8, 16, 24, 32…

These are in an A.P., having first term as 8 and common difference as 8.

Therefore, a = 8

d = 8

S₁₅ =?

= 960

Answer:

The odd numbers between 0 and 50 are

1, 3, 5, 7, 9 … 49

Therefore, it can be observed that these odd numbers are in an A.P.

a = 1

d = 2

l = 49

l = a + (n − 1) d

49 = 1 + (n − 1)2

48 = 2(n − 1)

n − 1 = 24

n = 25

= 625

Answer:

It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.

a = 200

d = 50

Penalty that has to be paid if he has delayed the work by 30 days = S₃₀

= 15 [400 + 1450]

= 15 (1850)

= 27750

Therefore, the contractor has to pay Rs 27750 as penalty.

Answer:

Let the cost of 1^st prize be P.

Cost of 2^nd prize = P − 20

And cost of 3^rd prize = P − 40

It can be observed that the cost of these prizes are in an A.P. having common difference as −20 and first term as P.

a = P

d = −20

Given that, S₇ = 700

a + 3(−20) = 100

a − 60 = 100

a = 160

Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

Answer:

It can be observed that the number of trees planted by the students is in an AP.

1, 2, 3, 4, 5………………..12

First term, a = 1

Common difference, d = 2 − 1 = 1

= 6 (2 + 11)

= 6 (13)

= 78

Therefore, number of trees planted by 1 section of the classes = 78

Number of trees planted by 3 sections of the classes = 3 × 78 = 234

Therefore, 234 trees will be planted by the students.

Answer:

Semi-perimeter of circle = πr

I₁ = π(0.5)

I₂ = π(1) = π cm

I₃ = π(1.5) =

Therefore, I₁, I₂, I₃ ,i.e. the lengths of the semi-circles are in an A.P.,

S₁₃ =?

We know that the sum of n terms of an a A.P. is given by

= 143

Therefore, the length of such spiral of thirteen consecutive semi-circles will be 143 cm.

Answer:

It can be observed that the numbers of logs in rows are in an A.P.

20, 19, 18…

For this A.P.,

a = 20

d = a₂ − a₁ = 19 − 20 = −1

Let a total of 200 logs be placed in n rows.

S_n = 200

400 = n (40 − n + 1)

400 = n (41 − n)

400 = 41n − n²

n² − 41n + 400 = 0

n² − 16n − 25n + 400 = 0

n (n − 16) −25 (n − 16) = 0

(n − 16) (n − 25) = 0

Either (n − 16) = 0 or n − 25 = 0

n = 16 or n = 25

a_n = a + (n − 1)d

a₁₆ = 20 + (16 − 1) (−1)

a₁₆ = 20 − 15

a₁₆ = 5

Similarly,

a₂₅ = 20 + (25 − 1) (−1)

a₂₅ = 20 − 24

= −4

Clearly, the number of logs in 16^th row is 5. However, the number of logs in 25^th row is negative, which is not possible.

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16^th row is 5.

Answer:

The distances of potatoes are as follows.

5, 8, 11, 14…

It can be observed that these distances are in A.P.

a = 5

d = 8 − 5 = 3

= 5[10 + 9 × 3]

= 5(10 + 27) = 5(37)

= 185

As every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be two times of it.

Therefore, total distance that the competitor will run = 2 × 185

= 370 m

Alternatively,

The distances of potatoes from the bucket are 5, 8, 11, 14…

Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. Therefore, distances to be run are

10, 16, 22, 28, 34,……….

a = 10

d = 16 − 10 = 6

S₁₀ =?

= 5[20 + 54]

= 5 (74)

= 370

Therefore, the competitor will run a total distance of 370 m.

Answer:

Given A.P. is 121, 117, 113 …

a = 121

d = 117 − 121 = −4

a_n = a + (n − 1) d

= 121 + (n − 1) (−4)

= 121 − 4n + 4

= 125 − 4n

We have to find the first negative term of this A.P.

Therefore, 32^nd term will be the first negative term of this A.P.

Answer:

We know that,

a_n = a + (n − 1) d

a₃ = a + (3 − 1) d

a₃ = a + 2d

Similarly, a₇ = a + 6d

Given that, a₃ + a₇ = 6

(a + 2d) + (a + 6d) = 6

2a + 8d = 6

a + 4d = 3

a = 3 − 4d (i)

Also, it is given that (a₃) × (a₇) = 8

(a + 2d) × (a + 6d) = 8

From equation (i),

From equation (i),

Answer:

It is given that the rungs are 25 cm apart and the top and bottom rungs arem apart.

∴ Total number of rungs

Now, as the lengths of the rungs decrease uniformly, they will be in an A.P.

The length of the wood required for the rungs equals the sum of all the terms of this A.P.

First term, a = 45

Last term, l = 25

n = 11

Therefore, the length of the wood required for the rungs is 385 cm.

Answer:

The number of houses was

1, 2, 3 … 49

It can be observed that the number of houses are in an A.P. having a as 1 and d also as 1.

Let us assume that the number of x^th house was like this.

We know that,

Sum of n terms in an A.P.

Sum of number of houses preceding x^th house = S_{x − 1}

Sum of number of houses following x^th house = S₄₉ − S_x

It is given that these sums are equal to each other.

However, the house numbers are positive integers.

The value of x will be 35 only.

Therefore, house number 35 is such that the sum of the numbers of houses preceding the house numbered 35 is equal to the sum of the numbers of the houses following it.

Answer:

From the figure, it can be observed that

1^st step is m wide,

2^nd step is 1 m wide,

3^rd step is m wide.

Therefore, the width of each step is increasing by m each time whereas their height m and length 50 m remains the same.

Therefore, the widths of these steps are

Volume of concrete in 1^st step

Volume of concrete in 2^nd step

Volume of concrete in 3^rd step

It can be observed that the volumes of concrete in these steps are in an A.P.

Volume of concrete required to build the terrace is 750 m³.