R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 5 - Arithmetic Progressions

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Chapter 11 - Arithmetic Progressions Ex. 5B

Solution 9

Let the required numbers be (a - d), a, (a + d)

Sum of these numbers = (a - d) + a + (a + d) = 3a

sum of these squares =

Sum of three numbers = 21, sum of squares of these numbers = 165

3a = 21

a = 7

Thus, a = 7 and d =

Hence, the required numbers are (4, 7, 10) or (10, 7, 4)

Solution 1

Solution 2

Solution 3

Solution 4

If are consecutive terms of an AP, then

Solution 5

Solution 6

Let the required numbers be (a - d), a and (a + d)

Sum of these numbers = (a - d) + a + (a + d) = 3a

Product of these numbers = (a - d) × a × (a + d) =

But sum = 15 and product = 80


Hence, the required numbers are (2, 5, 8)

Solution 7

Let the required numbers be (a - d), a, (a + d)

Sum of these number = (a - d) + a + (a + d) = 3a

Product of these numbers = (a - d) a (a + d)

                                    

But,sum = 3 and product = - 35

Thus, a = 1 and d = 6

Hence, the required numbers are (-5, 1, 7)

Solution 8

Let the required number be (a - d), a and (a + d)

Sum of these numbers = (a - d) + a + (a + d) = 3a

Product of these numbers = (a - d) x a x (a + d)

But sum = 24 and product = 440

Thus, a = 8 and d = 3

Hence the required numbers are (5, 8, 11)

Solution 10

Let the required angles be (a - 3d)°, (a - d) °, (a + d) ° and (a + 3d) °

Common difference = (a - d) - (a- 3d) = a - d - a + 3d = 2d

Common difference = 10°

2d = 10° = d = 5°

Sum of four angles of quadrilateral = 360°


First angle = (a - 3d)° = (90 - 3 × 5) ° = 75°

Second angle = (a - d)° = (90 - 5) ° = 85°

Third angle = (a + d)° = (90 + 5°) = 95°

Fourth angle = (a + 3d)° = (90 + 3 × 5)° = 105°

Solution 11

Let the required number be (a - 3d), (a - d), (a + d) and (a + 3d)

Sum of these numbers = (a - 3d) + (a - d)+ (a + d) + (a + 3d)

4a = 28 a = 7

Sum of the squares of these numbers

 

Hence, the required numbers (4, 6, 8, 10)

Solution 12

Solution 13

Chapter 11 - Arithmetic Progressions Ex. 5C

Solution 26

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(v)

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 3

Solution 4

Solution 6(i)

It is given that -----(1)

Now, 20th term

=(sum of first 20 term) - (sum of first 19 terms)

Putting = 20 in (1) we get

Putting n= 19 in (1), we get


Hence, the 20thterm is 99

Solution 5

 

(i)The nth term is given by


(ii)Putting n = 1 in (1) , we get


(iii)Putting n = 2 in (1), we get = 8

Solution 6(ii)

Solution 8

Here a = 21, d = (18 - 21) = -3

Let the required number of terms be n, then


sum of first 15 terms = 0

Solution 9

Solution 10

Solution 11

Solution 12

The term AP is

Solution 13

Solution 14

Solution 15

Solution 16

All numbers between 300 and 700 that are multiples of 9 are 306, 315, 324, 333, ..., 693

This is an AP in which a = 306, d = (315 - 306) = 9, l = 693

Let the number of these terms be n, then

Solution 17


Thus, common difference = 3

Solution 18

The given AP is

Common difference d =

Term next to 

Solution 19

The given AP is

First term

Common difference d =

Sum of n terms =

Solution 20

Solution 21

First term 'a' of an AP = 2

The last term l = 29

 

common difference = 3

Solution 22

Solution 23

Solution 24

Solution 25

First term of an AP, a = 22

Last term = nth term = - 11

 

Thus, n = 12, d = -3

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34(i)

Solution 34(ii)

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47

Solution 48

Chapter 11 - Arithmetic Progressions MCQ

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Chapter 5 - Arithmetic Progressions Ex. 5A

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(v)

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 2(iv)

Solution 2(v)

Solution 3(i)

The given AP is

First term = 6, common difference =

                                               

a = 6, d =

The nth term is given by

Hence, 37th term is 69

Solution 3(ii)

The given AP is

The first term = 5,

common difference =

a = 5,

The nth term is given by

Hence the 25th term is - 7

Solution 5(i)

Solution 5(ii)

Solution 6

 

(i)First term = -6

(ii)Common difference

(iii)16th term = where a = -6 and d = 4

       = (-6 + 15 4) = 54

Solution 7

In the given AP, we have a = 6 and d = (10 - 6) = 4

Suppose there are n terms in the given AP, then


Hence there are 43 terms in the given AP

Solution 8

In the given AP we have a = 41 and d = 38 - 41 = - 3

Suppose there are n terms in AP, then

 

Hence there are 12 terms in the given AP

Solution 9

Solution 10

In the given AP, we have a = 3 and d = 8 - 3 = 5

Suppose there are n terms in given AP, then

Hence, the 18th term of given AP is 88

Solution 11

In the given AP, we have a = 72 and d = 68 - 72 = - 4

Suppose there are n terms in given AP, we have

 

Hence, the 19th term in the given AP is 0


Solution 12

In the given AP, we have

Suppose there are n terms in given AP, we have

Then,

 

Thus, 14th term in the given AP is 3

Solution 13

Solution 15

The given AP is 5, 15, 25....

a = 5, d = 15 - 5 = 10

Thus, the required term is 44th

Solution 16

In the given AP let the first term = a,

And common difference = d


So the required AP is 7, 12, 17, 22....

Solution 17

Solution 18

Solution 19

Solution 20

Here a = 7, d = (10 - 7) = 3, l = 184

And n = 8

Hence, the 8th term from the end is 163

Solution 21

Here a = 17, d = (14 - 17) = -3, l = -40

And n = 6

Now, nth term from the end = [l - (n - 1)d]

                                      

Hence, the 6th term from the end is - 25

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

In the given AP, let the first term = a common difference = d

So the required AP is 8, 6, 4, 2, 0......

Solution 27

Let the first term of given AP = a and common difference = d

Hence 25th term is triple its 11th term

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Let a be the first term and d be the common difference

Solution 34

Solution 35

Solution 36

First AP is 63, 65, 67....

First term = 63, common difference = 65 - 63 = 2

nthterm = 63 + (n - 1) 2 = 63 + 2n - 2 = 2n + 61

Second AP is 3, 10, 17 ....

First term = 3, common difference = 10 - 3 = 7

nth term = 3 + (n - 1) 7 = 3 + 7n - 7 = 7n - 4

The two nth terms are equal

2n + 61 = 7n - 4 or 5n = 61 + 4 = 65

Solution 37

Solution 38

Solution 39

Solution 40

Let a be the first term and d be the common difference

pth term = a +(p - 1)d = q(given)-----(1)

qth term = a +(q - 1) d = p(given)-----(2)

subtracting (2) from (1)

(p - q)d = q - p

(p - q)d = -(p - q)

d = -1

Putting d = -1 in (1)

Solution 41

Let a be the first term and d be the common difference

nth term from the beginning = a + (n - 1)d-----(1)

nth term from end= l - (n - 1)d ----(2)

adding (1) and (2),

sum of the nth term from the beginning and nth term from the end = [a + (n - 1)d] + [l - (n - 1)d] = a + l

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Number of rose plants in first, second, third rows.... are 43, 41, 39... respectively.

There are 11 rose plants in the last row

So, it is an AP . viz. 43, 41, 39 .... 11

a = 43, d = 41 - 43 = -2, l = 11

Let nth term be the last term

Hence, there are 17 rows in the flower bed.

Solution 47

Chapter 5 - Arithmetic Progressions Ex. 5D

Solution 1

 

Solution 2

 

Solution 3

 

Solution 4

 

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

All three digit natural numbers divisible by 13 are 104, 117, 130, 143,..., 988

This is an AP in which a = 104, d = (117 - 104) = 13, l = 988

Solution 13

First 15 multiples of 8 are 8, 16, 24, ... to 15th term

Solution 14

Odd natural numbers between 0 and 50 are 1, 3, 5, ... 49

a = 1, d = 3 - 1= 2, l = 49

Let the number of terms be n

Solution 15

First 100 even natural numbers divisible by 5 are

10, 20, 30, ... to 100 term

First term of AP = 10

Common difference d = 20 - 10 = 10

Number of terms = n = 100

Solution 16

The given AP is 21, 18, 15, ....

First term = 21, common difference = 18 - 21= - 3

Let nth term be zero

a + (n - 1)d = 0or 21 + (n - 1)(-3) = 0

21 - 3n + 3 = 0

3n = 24

or 

Hence, 8th term of given series is 0

Solution 17

Sum of n natural numbers = 1 + 2 + 3 + ... + n

Here a = 1, d = 2 - 1 = 1

Solution 18

Sum of even natural numbers = 2 + 4 + 6 + ... to n terms

a = 2, d = 4 - 2 = 2

Solution 19

First term of AP = a = p

Common difference = d = q

nth term = a + (n 1)d

10th term = p + (10 1)q

               = p + 9q

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Other Chapters for CBSE Class 10 Maths

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Linear equations in two variables Chapter 4- Quadratic Equations Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- T-Ratios of Some Particular Angles Chapter 12- Trigonometric Ratios of Complementary Angles Chapter 13- Trigonometric Identities Chapter 14- Height and Distance Chapter 15- Perimeter and Areas of Plane Figures Chapter 16- Areas of Circle, Sector and Segment Chapter 17- Volume and Surface Areas of Solids Chapter 18- Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Chapter 19- Probability

R S AGGARWAL AND V AGGARWAL Solutions for CBSE Class 10 Subjects