R S Aggarwal Solutions for Class 10 Maths Chapter 5 Arithmetic Progression

R S Aggarwal Solutions for Class 10 Chapter 5 Arithmetic Progression are available here. Chapter 5 of Class 10 gives the students an overview of the different problems related to the topic. All questions are prepared by subject experts at BYJU’S, based on the CBSE Syllabus. Students of Class 10 are often advised to solve all R S Aggarwal Solutions for Maths to sharpen their skills. Students can avail the R S Aggarwal Solutions and download the pdf for free to practice offline.

 

Access Solutions to Maths R S Aggarwal Chapter 5 Arithmetic Progression

Get detailed solutions for all the questions listed under the below exercises:

Exercise 5A Solutions: 30 Questions (Short Answers)

Exercise 5B Solutions: 16 Questions (Short Answers)

Exercise 5C Solutions : 15 Questions (Short Answers)

Exercise 5D Solutions: 20 Questions (Long Answers)

Exercise 5A Page No: 257

Question 1:

 

 

Solution:

(i) 9, 15, 21, 27, …

Here, 15 – 9 = 21 – 15 = 27 – 21 = 6 (which is constant)

Common difference is 6

Or d = 6

Next term = 27 + d = 27 + 6 = 33

(ii) 11, 6, 1, -4, …

Here, 6 – 11 = 1 – 6 = -5 -4 – 1 = -5 (which is constant)

d (common difference) = -5

Next term = -4 – 5 = -9

(iii) -1, -5 /6 , -2 /3 , -1 /2 , …………

-5/6 – (-1) = 1/6 and

-2/3 – (-5/6) = 1/6

d (common difference) = 1/6

Next term = -1/2 + 1/6 = -1/3

(iv) √2, √8, √18, √32, ……..

√8 – √2 = 2√2 – √2 = √2

√32 – √18 = 4√2 – 3√2 = √2

d (common difference) = √2

Next term = √32 + √2 = 4√2 + √2 = 5√2 = √50

(v) √20, √45, √80, √125, ……

√20, √45, √80, √125, ……

√45 – √20 = 3√5 – 2√5 = √5

√125 – √80 = 5√5 – 4√5 = √5

d (common difference) = √5

Next term = √125 + √5 = 5√5 + √5 = 6√5 or √180

Question 2:

 

 

Solution:

(i)Given: AP is 9, 13, 17, 21, ……

Here, first term = a = 9

Common difference = d = 13 – 9 = 4

an = a + (n-1)d

a20= 9 + (20-1)4

= 85

(ii) the 35th term of the AP 20, 17, 14, 11, ……

Given: AP is 20, 17, 14, 11, ……

Here, first term = a = 20

Common difference = d = 17 – 20 = -3

n = 35

an = a + (n-1)d

a35 = 20 + (35-1)(-3)

= -82

(iii) the 18th term of the AP √2, √18 , √50, √98, ………

Given: AP is √2, √18 , √50, √98, ………

or √2, 3√2 , 5√2, 7√2,….

Here, first term = a = √2

Common difference = d = 3√2 – √2 = 2√2

n = 18

an = a + (n-1)d

a18 = √2 + 34√2

= 35√2

(iv) the 9th term of the AP 3 /4 , 5 /4 , 7 /4 , 9 /4 , …….

Given: AP is 3 /4 , 5 /4 , 7 /4 , 9 /4 , …….

Here, first term = a = 3/4

Common difference = d = 5/4 – 3/4 = 1/2

n = 9

an = a + (n-1)d

a9 = 3/4 + (9 -1)1/2

= 19/4

(v) the 15th term of the AP -40, -15, 10, 35, ……

Given: AP is -40, -15, 10, 35, ……

Here, first term = a = -40

Common difference = d = -15 – (-40) = -15 + 40 = 25

n = 15

an = a + (n-1)d

a15 = -40 + (15 – 1)25

= 310

Question 3:

 

 

Solution:

rs aggarwal class 10 chapter 5 ex a 1

n = 25

an = a + (n-1)d

a25 = 5 + (25-1)(-1/2)

= -7

Question 4:

 

 

Solution:

If 2p – 1, 3p + 1, 11 are terms in AP, then

a2 – a1 = a3 – a2 …..(1)

From given:

a1 = 2p – 1

a2 = 3p + 1

a3 = 11

From (1), we get

(3p + 1) – (2p – 1) = 11 – (3p + 1)

3p + 1 – 2p + 1 = 11 – 3p – 1

p + 2 = 10 – 3p

4p = 8

p = 2

For p = 2, these terms are in AP.

Question 5:

 

 

Solution:

(i)AP is 5, 11, 17, 23, ……

Here, first term = a = 5

Common difference = d = 11 – 5 = 6

Now,

an = a (n – 1)d

= 5 + (n – 1) 6

= 5 + 6n – 6

= (6n – 1)

(ii) AP is 16, 9, 2, -5, ……

Here, first term = a = 16

Common difference = d = 9 – 16 = -7

an = a + (n – 1)d

= 16 + (n – 1) (-7)

= 16 – 7n + 7

= (23 – 7n)

Question 6:

 

 

Solution:

nth term of AP is 4n – 10 (Given)

Putting n = 1, 2, 3, 4, …, we get

At n = 1: 4n – 10 = 4 x 1 – 10 = 4 – 10 = -6

At n = 2: 4n – 10 = 4 x 2 – 10 = 8 – 10 = -2

At n = 3: 4n – 10 = 4 x 3 – 10 = 12 – 10 = 2

At n = 1: 4n – 10 = 4 x 4 – 10 = 16 – 10 = 6

We see that -6, -2, 2, 6,… are in AP

(i) first term = -6

(ii) Common difference = -2 – (-6) = 4

(iii) 16th term:

Using formula: an = a + (n – 1)d

Here n = 16

a16 = -6 + (16 – 1)4 = 54

Question 7:

 

 

Solution:

Given: AP is 6, 10, 14, 18,…, 174

Here, first term = a = 6

Common difference = d= 10 – 6 = 4

To find: the number of terms (n)

Last term = a + (n – 1)d

174 = 6 + (n – 1) 4

174 – 6 = (n – 1) 4

n – 1 = 168 /4 = 42

n = 42 + 1 = 43

There are 43 terms.

Question 8:

 

 

Solution:

Given: AP is 41, 38, 35,…, 8

Here, first term = a = 41

Last term = 8

Common difference = d = 38 – 41 = -3

To find: the number of terms (n)

Last term = a + (n – 1)d

8 = 41 + (n – 1)(-3)

8 – 41 = (n – 1)(-3)

n – 1 = 11

n = 11 + 1 = 12

There are 12 terms.

Question 9:

 

 

Solution:

Given: AP is 18, 31/2 , 13, …, -47

Here, first term = a = 18

Last term = -47

Common difference = d = -5/2

To find: the number of terms (n)

Last term = a + (n – 1)d

-47 = 18 + (n – 1)(-5/2)

-47 – 18 = (n – 1)(-5/2)

n = 27

There are 27 terms.

Question 10:

 

 

Solution:

Let nth term is 88.

AP is 3, 8, 13, 18, …

Here,

First term = a = 3

Common difference = d = 8 – 3 = 5

nth term of AP is an = a + (n – 1) d

Now,

88 = 3 + (n – 1)(5)

88 – 3 = (n – 1) x 5

n – 1 = 88 /5

or n = 17 + 1 = 18

Therefore: 88 is the 18th term.

Question 11:

 

Solution:

AP is 72, 68, 64, 60, …..

Let nth term is 0.

Here,

First term = a = 72

Common difference = d = 68 – 72 = -4

an = a + (n – 1)d

0 = 72 + (n – 1)(-4)

-72 = -4(n – 1)

n – 1 = 18

n = 18 + 1 = 19

Therefore: 0 is the 19th term.

Question 12:

 

 

Solution:

rs aggarwal class 10 chapter 5 ex a 2

Here,

First term = a = 5/6

Common difference = d = 1 – 5/6 = 1/6

Now: an = a + (n – 1)d

3 = 5/6 + (n – 1)1/6

Let nth term is 3

Now, an = a + (n-1)d

3 = 5/6 + (n-1)1/6

n -1 = 13

n = 13 + 1 = 14

Therefore, 3 is the 14th term.

Question 13:

 

 

Solution:

AP is 21, 18, 15, ……

Let nth term -81

Here, a = 21, d = 18 – 21 = -3

an = a + (n – 1)d

-81 = 21 + (n – 1)(-3)

-81 – 21 = (n – 1)(-3)

-102 = (n – 1)(-3)

n = 34 + 1 = 35

Therefore, -81 is the 35th term

Question 14:

 

 

Solution:

Given AP is 3, 8, 13, 18,…

First term = a = 3

Common difference = d = 8 – 3 = 5

And n = 20 and a20 be the 20th term, then

a20 = a + (n – 1)d

= 3 + (20 – 1) 5

= 3 + 95

= 98

The required term = 98 + 55 = 153

Now, 153 be the nth term, then

an = a + (n – 1)d

153 = 3 + (n – 1) x 5

153 – 3 = 5(n – 1)

150 = 5(n – 1)

n – 1 = 30

n = 31

Required term will be 31st term.

Question 15:

 

 

Solution:

AP is 5, 15, 25,…

First term = a = 5

Common difference = d = 15 – 5 = 10

Find 31st term:

a31 = a + (n – 1)d

= 5 + (31 – 1) 10

= 5 + 30 x 10

= 305

Required term = 305 + 130 = 435

Now, say 435 be the nth term, then

an = a + (n – 1)d

435 = 5 + (n – 1)10

435 – 5 = (n – 1)10

n – 1 = 43

n = 44

The required term will be 44th term.

Question 16:

 

 

Solution:

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

rs aggarwal class 10 chapter 5 ex a 3

Question 17:

 

 

rs aggarwal class 10 chapter 5 ex a 4

a16 = 6 + (16 – 1)7 = 6 + 105 = 111

Therefore, midterm of the AP id 111.

Question 18:

 

 

Solution:

AP is 10, 7, 4, …..…, (-62)

a = 10,

d = 7 – 10 = -3, and

l = -62

Now, an = l = a + (n – 1)d

-62 = 10 + (n – 1) x (-3)

-62 – 10 = -3(n- 1)

-72 = -3(n – 1)

Or n = 24 + 1 = 25

Middle term = (25 + 1) /2 th = 13th term

Find the 13th term using formula, we get

a13 = 10 + (13 – 1)(-3) = 10 – 36 = -26

Question 19:

 

 

Solution:

Given AP is -4 /3 , -1, -2 /3 , …, 13 /3

rs aggarwal class 10 chapter 5 ex a 5

rs aggarwal class 10 chapter 5 ex a 6

Middle terms will be: (18/2)th + (18/2 + 1)th

= 9th + 10th term

Now,

a9 + a10 = a + 8d + a + 9d

= 2a + 17d

= 2(-4/3) + 17(1/3)

= 3

Question 20:

 

Solution:

Given: AP is 7, 10, 13,…, 184

a = 7, d = 10 – 7 = 3 and l = 184

nth term from the end = l – (n – 1)d

Now,

8th term from the end be

184 – (8 – 1)3 = 184 – 21 = 163

Question 21:

 

 

Solution:

Given: AP is 17, 14, 11, …,(-40)

a = 17, d = 14 – 17 = -3, l = -40

6th term from the end = l – (n – 1)d

= -40 – (6 – 1) (-3)

= -40 – (5 x (-3))

= -40 + 15

= -25

Question 22:

 

 

Solution:

Given AP is 3, 7, 11, 15, …

a = 3, d = 7 – 3 = 4

Let 184 be the nth term of the AP

an = a + (n – 1)d

184 = 3 + (n – 1) x 4

184 – 3 = (n – 1) x 4

181 /4 = n – 1

n = 181 /4 + 1 = 185/4 (Which is in fraction)

Therefore, 184 is not a term of the given AP.

Question 23:

 

 

Solution:

Given AP is AP 11, 8, 5, 2,…

Here a = 11, d = 8-11= -3

Let -150 be the nth term of the AP

an = a + (n – 1)d

-150 = 11 +(n-1)(-3)

or n = 164/3

Which is a fraction.

Therefore, -150 is not a term of the given AP.

Question 24:

 

 

Solution:

Let nth of the AP 121, 117, 113,… is negative. Let Tn be the nth term then

rs aggarwal class 10 chapter 5 ex a 7

Therefore, 32nd term will be the 1st negative term.

Question 25:

 

 

Solution:

rs aggarwal class 10 chapter 5 ex a 8

a = 20, d = -3/4

Let nth term be the 1st negative term of the AP

an < 0

an = a + (n – 1)d

rs aggarwal class 10 chapter 5 ex a 9

Therefore, 28th term will be the 1st negative term.

 

 

Solution:

Let us say a be the first term and d be the common difference of an AP

an = a + (n – 1)d

a7 = a + (7 – 1)d

= a + 6d = -4 …………(1)

And a13 = a + 12d = -16 …..…..(2)

Subtracting equation (1) from (2), we get

6d = -16 – (-4) = -12

From (1), a + 6d = -4

a + (-12) = -4

a = -4 + 12 = 8

a = 8, d = -2

AP will be 8, 6, 4, 2, 0, ……

Question 27:

 

 

Solution:

Let a be the first term and d be the common difference of an AP.

a4 = a + (n- 1)d

= a + (4 – 1)d

= a + 3d

Since 4th term of an AP is zero.

a + 3d = 0

or a = -3d ….(1)

Similarly,

a25 = a + 24d = -3d + 24d = 21d …(2)

a11 = a + 10d = -3d + 10d = 7d ….(3)

From (2) and (3), we have

a25 = 3 x a11

Hence proved.

Question 28:

 

Solution:

Sixth term of an AP is zero

that is a6 = 0

a + 5d = 0

a = -5 d

Now, a15 = a + (n – 1 )d

a + (15 – 1)d = -5d + 14d = 9d

and a33 = a + (n – 1 )d = a + (33 – 1)d = -5d + 32d = 27d

Now, a33 : a12

27d : 9d

3 : 1

Which shows that a33 = 3(a15)

Hence proved.

Question 29:

 

Solution:

Let a be the first term and d be the common difference of an AP.

an = a + (n – 1)d

a4 = a + (4 – 1)d = a + 3d

a + 3d = 11 ………(1)

Now, a5 = a + 4d and a7 = a + 6d

Now, a5 + a7 = a + 4d + a + 6d = 2a + 10d

2a + 10d = 34

a + 5d = 17 ……..(2)

Subtracting (1) from (2), we get

2d = 17 – 11 = 6

d = 3

The common difference = 3

Question 30:

 

 

Solution:

Let a be the first term and d be the common difference of an AP.

nth term = an = a + (n – 1)d

Given: 9th term of an AP is -32 and the sum of its 11th and 13th terms is -94

Now,

a9 = a + 8d = -32 …(1)

a11 = a + 10d

a13 = a + 12d

Sum of 11th and 13th terms:

a11 + a13 = a + 10d + a + 12d

-94 = 2a + 22d

or a + 11d =-47 …(2)

Subtracting (1) from (2) , we have

3d = -47 + 32 = -15

or d = -5

Common difference is -5.


Exercise 5B Page No: 264

Question 1:

 

Solution:

Given: (3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an AP.

So, (4k – 6) – (3k – 2) = (k + 2) – (4k – 6)

2(4k – 6) = (k + 2) + (3k – 2)

8k – 12 = 4k + 0

8k – 4k = 0 + 12

or k = 3

Question 2:

 

Solution:

Given: (5x + 2), (4x – 1) and (x + 2) are terms in AP.

So, d = (4x – 1) – (5x + 2) = (x + 2) – (4x – 1)

2(4x – 1) = (x + 2) + (5x + 2)

8x – 2 = 6x + 2 + 2

8x – 2 = 6x + 4

8x – 6x = 4 + 2

or x = 3

The value of x is 3.

Question 3:

 

Solution:

Given: (3y – 1), (3y + 5) and (5y + 1) are 3 consecutive terms of an AP.

So, (3y + 5) – (3y – 1) – (5y + 1) – (3y + 5)

2(3y + 5) = 5y + 1 + 3y – 1

6y + 10 = 8y

8y – 6y = 10

2y = 10

Or y = 5

The value of y is 5.

Question 4:

 

Solution:

Given: (x + 2), 2x, (2x + 3) are three consecutive terms of an AP.

So, 2x – (x + 2) = (2x + 3) – 2x

2x – x – 2 = 2x + 3 – 2x

x – 2 = 3

x = 2 + 3 = 5

The value of x is 5.

Question 5:

 

Solution:

Assume that (a – b)², (a² + b²) and (a + b)² are in AP.

So, (a² + b²) – (a – b)² = (a + b)² – (a² + b²)

(a² + b²) – (a² + b² – 2ab) = a² + b² + 2ab – a² – b²

2ab = 2ab

Which is true.

Hence given terms are in AP.

Question 6:

 

 

Solution:

Let a – d, a, a + d are three numbers in AP.

So, their sum = a – d + a + a + d = 15

3a = 15

or a = 5

Again,

Their Product = (a – d) x a x (a + d) = 80

a(a² – d²) = 80

5(5² – d²) = 80

25 – d² = 16

d² = 25 – 16 = 9 = (±3)²

or d = ±3

d = 3 or d = -2

We have 2 conditions here:

At a = 5, d = 3

Numbers are: 2, 5 and 8

At a = 5 and d = -3

Numbers are : 8, 5, 2

Question 7:

 

Solution:

Let a – d, a, a + d are three numbers in AP.

their sum = a – d + a + a + d = 3

3a = 3

or a = 1

Again,

Their Product = (a – d) x a x (a + d) = -35

a(a² – d²) = -35

(1² – d²) = -35

or d = ±6

d = 6 or d = -6

We have 2 conditions here:

At a = 1, d = 6

Numbers are: -5, 1 and 7

At a = 1 and d = -6

Numbers are : 7, 1, -5

Question 8:

 

Solution:

Let a – d, a, a + d are three numbers in AP.

their sum = a – d + a + a + d = 24

3a = 24

or a = 8

Again,

Their Product = (a – d) x a x (a + d) = 440

a(a² – d²) = 440

8(8² – d²) = 440

or d = ±3

Numbers are : (5, 8, 11) or (11, 8, 5)

Question 9:

 

 

Solution:

Let a – d, a, a + d are three numbers in AP.

their sum = a – d + a + a + d = 21

3a = 21

or a = 7

Again,

Sum of squares = (a – d)^2 + a^2 +(a + d)^2 = 165

rs aggarwal class 10 chapter 5 ex b 1

Numbers are : (4, 7, 10) or (10, 7, 4)

Question 10:

 

Solution:

Sum of angles of a quadrilateral = 360°

Common difference = 10 = d (say)

If the first number be a, then the next four numbers will be

a, a + 10, a + 20, a + 30

As per definition:

a + a + 10 + a + 20 + a + 30 = 360°

4a + 60 = 360

4a = 300

or a = 75°

Other angles:

a + 10 = 75 + 10 = 85

a + 20 = 75 + 20 = 95

a + 30 = 75 + 30 = 105

Therefore, Angles are 75°, 85°, 95°, 105°

Question 11:

 

Solution:

Let a – 3d, a – d, a + d, a + 3d are the four numbers in AP.

Their sum = a – 3d + a – d + a + d + a + 3d = 28

Sum of their square = (a – 3d)^2 + (a – d)^2 + (a + d)^2 + (a + 3d)^2 = 216

rs aggarwal class 10 chapter 5 ex b 2

Numbers will be: (4, 6, 8,10) or ( 10, 8, 6, 4)

Question 12:

Solution:

Let a – 3d, a – d, a + d, a + 3d are the four numbers in AP.

Therefore:

rs aggarwal class 10 chapter 5 ex b 3

d = ±2

four parts are: a – 6, a – 2, a+2 amd a + 6

which implies,

Number are: (2, 6, 8, 10, 14) or (14, 10, 6, 2)

Question 13:

Solution:

Let a – d, a, a + d are the three terms in AP

So,

Sum = a – d + a + a + d = 48

3a = 48

Or a = 16

And,

a(a – d) = 4 (a + d) + 12

16 (16 – d) = 4(16 + d) + 12

256 – 16d = 64 + 4d + 12 = 4d + 76

256 – 76 = 4d + 16d

180 = 20d

d = 9

Which implies:

Numbers are : (7, 16, 25)


Exercise 5C Page No: 281

Question 1:

 

Solution:

Sum of n terms of AP formula:

rs aggarwal class 10 chapter 5 ex c 1……(1)

Where

First term = a

Common difference = d

Number of terms = n

(i) 2, 7, 12, 17,….. to 19 terms.

a = 2

d = 7 – 2 = 5

Using (1)

S19 = 19/2(2(2) + (19 – 1)5)

= (19)(4 + 90)/2

= (19 × 94)/2

= 893

Sum of 19 terms of this AP is 893.

(ii) 9, 7, 5, 3,…..to 14 terms.

a = 9

d = 7 – 9 = -2

Using (1)

S14 = 14/2 [2(9) + (14 – 1)(-2)]

= (7)(18 – 26)

= – 56

Sum of 14 terms of this AP is – 56.

(iii) -37, -33, -29,….12 terms.

a = -37

d = (-33) – (-37) = 4

Using (1)

S12 = 12/2 [2(-37) + (12 – 1)(4)]

= (6)(-74 + 44)

= 6 × (-30)

= – 180

Sum of 12 terms of this AP is – 180.

(iv) 1/15 ,1/12,1/10,…to 11 terms

a = 1/15

d = (1/12) – (1/15) = 1/60

Using (1)

S11 = 11/2 [2(1/15) + (11 – 1)(1/60)]

= (11/2) × [(2/15) + (1/6)]

=33/20

Sum of 11 terms of this AP is 33/20.

(v) 0.6, 1.7, 2.8,…. to 100 terms.

a = 0.6

d = 1.7 – 0.6 = 1.1

Using (1)

S100 = 100/2 [2(0.6) + (100 – 1)(1.1)]

= (50) × [1.2 + (99 × 1.1) ]

= 50 × 110.1

= 5505

Sum of 100 terms of this AP is 5505.

Question 2:

 

Solution:

(i) 7 + 10 1/2 + 14 + … + 84

First term = a = 7

Common difference = d = (21/2) – 7 = (7/2)

Last term = l = 84

Now, using formula:

84 = a + (n – 1)d

84 = 7 + (n – 1)(7/2)

77 = (n – 1)(7/2)

154 = 7n – 7

7n = 161

n = 23

Thus, there are 23 terms in AP.

Now,

Find Sum of these 23 terms:

S23 = 23/2 [2(7) + (23 – 1)(7/2)]

= (23/2) [14 + (22)(7/2) ]

= (23/2) [91]

= 2093/2

Sum of 23 terms of this AP is 2093/2.

(ii) 34 + 32 + 30 + … + 10

First term = a = 34

Common difference = d = 34 – 32 = – 2

Last term = l = 10

Now, using formula:

10 = a + (n – 1)d

10 = 34 + (n – 1)(-2)

10 – 34 = (n – 1)(-2)

n = 13

Thus, there are 13 terms in AP.

Now,

Find Sum of these 13 terms:

S13 = 13/2 [2(34) + (13 – 1)(-2)]

= (13/2) [68 + (12)(-2)]

= (13/2) x 44

= 286

Sum of 13 terms of this AP is 286.

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

First term = a = -5

Common difference = d = – 8 – (-5) = – 3

Last term = l = -230

Now, using formula:

-230 = a + (n – 1)d

-230 = -5 + (n – 1)(-3)

– 230 + 5 = (n – 1)(-3)

n = 76

Thus, there are 76 terms in AP.

Now,

Find Sum of these 76 terms:

S76 = 76/2 [2(-5) + (76 – 1)(-3)]

= 38 × [(-10) + (75)(-3)]

= – 8930

Sum of 76 terms of this AP is -8930.

(iv) 5 + (-41) + 9 + (-39) + 13 + (-37) + 17 + … + (-5) + 81 + (-3)

The given series is combination of two AP’s.

Let A1 = 5 + 9 +13 + ……+ 77 + 81

and A2 = -41 – 39 – 37 – ……(-3)

For A1:

First term = a = 5

Common difference = d = 2

Last term = l = 81

Now, using formula:

81 = 5 + (n – 1)4

n = 20

Thus, there are 20 terms in AP.

Now,

Find Sum of these 76 terms:

S20 = n/2 [a + l]

= 20/2 (5 + 81)

= 860

Sum of 20 terms of this AP is 860.

For A2:

First term = a = -41

Common difference = d = 2

Last term = l = -3

Now, using formula:

-3 = -41 + (n – 1)2

n = 20

Thus, there are 20 terms in AP.

Now,

Find Sum of these 76 terms:

S20 = n/2 [a + l]

= 20/2 (-41- 3)

=-440

Sum of 20 terms of this AP is -440.

Therefore, Sum of total terms: 860 – 440 = 420.

Question 3: 

 

Solution:

Given: an = 5 – 6n

Find some of the terms of AP:

Put n = 1, we get a1 = – 1 = first term

Put n = 2, we get a2 = – 7 = second term

Common difference = d = a2 – a1 = – 7 – (-1) = – 6

Sum of first n terms:

Sn= n/2 [2a + (n – 1)d]

= n/2[- 2 + (n – 1)(-6)]

= n(2 – 3n)

sum of first 20 terms:

S20 = 20/2[2(-1) + (20 – 1)(-6)]

= 10 [ – 2 – 114]

= – 1160

Sum of its first 20 terms of AP is -1160.

Question 4:

 

 

Solution:

Given: Sn = 3n² + 6n

rs aggarwal class 10 chapter 5 ex c 2

Question 5:

 

Solution:

Sn = 3n² – n

S1 = 3(1)² – 1 = 3 – 1 = 2

S2 = 3(2)² – 2 = 12 – 2 = 10

a1 = 2

a2 = 10 – 2 = 8

(i) an = a + (n – 1) d

= 2 + (n – 1) x 6

= 2 + 6n – 6

= 6n – 4

(ii) First term = 2

(iii) Common difference = 8 – 2 = 6

Question 6: 

 

Solution:

rs aggarwal class 10 chapter 5 ex c 3

rs aggarwal class 10 chapter 5 ex c 4

(ii)

rs aggarwal class 10 chapter 5 ex c 5

Question 7: 

 

Solution:

Let a be first term and d be the common difference of an AP.

mth term = 1/n

so,

am = a + (m – 1) d = 1/n ……….(1)

nth term = 1/m

an = a + (n – 1) d = 1/m ……….(2)

Subtract (2) from (1)

rs aggarwal class 10 chapter 5 ex c 6

Sum of first mn terms:

Smn = mn/2 [2(1/mn) + (mn-1) (1/mn)]

= mn/2 [1/mn + 1]

= (1+mn)/2

Question 8:

 

 

Solution:

AP is 21, 18, 15,…

a = 21,

d = 18 – 21 = -3

Sum of terms = Sn = 0

(n/2) [2a + (n – 1)d] = 0

(n/2) [2(21) + (n – 1)(-3)] = 0

(n/2) [45 – 3n] = 0

[45 – 3n] = 0

n = 15 (number of terms)

Thus, 15 terms of the given AP sums to zero.

Question 9: 

 

Solution:

AP is 9, 17, 25,…

a = 9, d = 17 – 9 = 8

Sum of terms = Sn = 636

(n/2) [2a + (n – 1)d] = 636

(n/2) [2(9) + (n – 1)(8)] = 636

(n/2)[10 + 8n] = 636

4n2 + 5n – 636 = 0 (which is a quadratic equation)

(n- 12)(4n + 53) = 0

Either (n- 12) = 0 or (4n + 53) = 0

n = 12 or n = – 53/4

Since n can’t be negative and fraction, so

n = 12

Number of terms = 12 terms.

Question 10: 

 

Solution:

AP is 63, 60, 57, 54,…

Here, a = 63, d = 60 – 63 = -3 and sum = Sn = 693

rs aggarwal class 10 chapter 5 ex c 7

Which is a quadratic equation

rs aggarwal class 10 chapter 5 ex c 8

Which shows that, 22th term of AP is zero.

Number of terms are 21 or 22. So there will be no effect on the sum.

Question 11:

 

 

Solution:

Here, a = 20, d = -2/3 and sum = sn = 300 (for n number of terms)

rs aggarwal class 10 chapter 5 ex c 9

rs aggarwal class 10 chapter 5 ex c 10

n(n-25) – 36(n-25) = 0

(n – 25)(n – 36) = 0

Either (n – 25) = 0 or (n – 36) = 0

n = 25 or n = 36

For n = 25

rs aggarwal class 10 chapter 5 ex c 11

= 300

Which is true.

Result is true for both the values of n

So both the numbers are correct.

Therefore, Sum of 11 terms is zero. (36-25 = 11)

Question 12:

Solution:

Odd numbers between 0 and 50 are 1, 3, 5, 7, 9, …, 49

Here, a = 1, d = 3 – 1 = 2, l = 49

rs aggarwal class 10 chapter 5 ex c 12

Sum of odd numbers:

= n/2(a + l)

= 25/2 (l + 49)

= 25/2(50)

= 625

Question 13: 

 

Solution:

Natural numbers between 200 and 400 which are divisible by 7 are 203, 210, 217,……,399.

Here,AP is 203, 210, 217,……,399

First term = a = 203,

Common difference = d = 7 and

Last term = l = 399

We know that, l = a + (n-1)d

rs aggarwal class 10 chapter 5 ex c 13

Sum of all 29 terms is 8729.

Question 14: 

 

Solution:

First forty positive integers which are divisible by 6 are

6, 12, 18, 24,…… to 40 terms

Here, a = 6, d = 12 – 6 = 6, and n = 40.

Sum of 40 terms:

rs aggarwal class 10 chapter 5 ex c 14

Question 15:

Solution:

First 15 multiples of 8 are as given below:

8, 16, 24, 32,….. to 15 terms

Here, a = 8, d = 16 – 8 = 8, and n = 15

Sum of 15 terms:

rs aggarwal class 10 chapter 5 ex c 15

Question 16: 

 

Solution:

Multiples of 9 lying between 300 and 700 = 306, 315, 324, 333, …, 693

Here, a = 306, d = 9, and l = 693

We know that, l = a + (n-1)d

rs aggarwal class 10 chapter 5 ex c 16

There are 44 terms.

Find the sum of 44 terms:

rs aggarwal class 10 chapter 5 ex c 17

Question 17:

 

Solution:

3-digit natural numbers: 100, 101, 102,……., 999.

3-digit natural numbers divisible by 13:

104, 117, 130,………, 988

Which is an AP.

Here, a = 104, d = 13, l = 988

l = a + (n – 1) d

988 = 104 + (n-1)13

n = 69

There are 69 terms.

Find the sum of 69 terms:

Sn = n/2(a + l)

= 69/2(104 + 988)

= 37674

Question 18: 

 

Solution:

Even natural numbers: 2, 4, 6, 8, 10, …

Even natural numbers divisible by 5: 10, 20, 30, 40, … to 100 terms

Here, a = 10, d = 20 – 10 = 10, and n = 100

Sn = n/2(2a + (n-1)d)

= 100/2 [2 x 10 + (100-1)10]

= 50500

Question 19:

 

(4 – 1/n) + (4-2/n) + (4– 3/n) + ….

Solution:

Given sum can be written as (4 + 4 + 4 + 4+….) – (1/n, 2/n, 3/n, …..)

Now, We have two series:

First series: = 4 + 4 + 4 + …… up to n terms

= 4n

Second series: 1/n, 2/n, 3/n, …..

Here, first term = a = 1/n

Common difference = d = (2/n) – (1/n) = (1/n)

Sum of n terms formula:

Sn = n/2[2a + (n – 1)d]

Sum of n terms of second series:

Sn = n/2 [2(1/n) + (n – 1)(1/n)]

= n/2 [(2/n) + 1 – (1/n)]

= (n + 1)/2

Hence,

Sum of n terms of the given series = Sum of n terms of first series – Sum of n terms of second series

= 4n – (n + 1)/2

= (8n – n – 1)/2

= 1/2 (7n – 1)

Question 20: 

 

Solution:

Let a be the first term and d be the common difference of the AP.

rs aggarwal class 10 chapter 5 ex c 18

rs aggarwal class 10 chapter 5 ex c 19

Which implies: a = 1 and d = 5

Therefore, the AP is 1, 6, 11, 16,……

Question 21: 

 

Solution:

Let d be the common difference.

Given:

first term = a = 2

last term = l = 29

Sum of all the terms = Sn = 155

Sn = n/2[a + l]

155 = n/2[2 + 29]

n = 10

There are 10 terms in total.

Therefore, 29 is the 10th term of the AP.

Now, 29 = a + (10 – 1)d

29 = 2 + 9d

27 = 9d

d = 3

The common difference is 3.

Question 22:

Solution:

Let d be the common difference.

Given:

first term = a = -4

last term = l = 29

Sum of all the terms = Sn = 150

Sn = n/2[a + l]

150 = n/2[-4 + 29]

n = 12

There are 12 terms in total.

Therefore, 29 is the 12th term of the AP.

Now, 29 = -4 + (12 – 1)d

29 = -4 + 11d

d = 3

The common difference is 3.

Question 23:

 

Solution:

Let n be the total number of terms.

Given:

First term = a = 17

Last term = l = 350

Common difference = d = 9

l = a + (n-1)d

350 = 17 + (n-1)9

n = 38

Again,

Sn = n/2[a + l]

= 38/2[17 + 350]

= 6973

There are 38 terms in total and their sum is 6973.

Question 24:

Solution:

Let n be the total number of terms and d be the common difference.

Given:

first term = a = 5

last term = l = 45

Sum of all terms = Sn = 400

Sn = n/2[a + l]

400 = n/2[5 + 45]

n/2 = 400/50

n = 16

There are 16 terms in the AP.

Therefore, 45 is the 16th term of the AP.

45 = a + (16 – 1)d

45 = 5 + 15d

40 = 15d

15d = 40

d = 8/3

Common difference = d = 8/3

Common difference is 8/3 and the number of terms are 16.

Question 25:

 

Solution:

Let n be the total number of terms and d be the common difference.

Given:

first term = a = 22

nth term = -11

Sum of all terms = Sn = 66

Sn = n/2[a + l]

66 = n/2[22 + (-11)]

n = 12

There are 12 terms in the AP.

Since nth term is -11, so

an = a + (n – 1)d

-11 = 22 + (12-1)d

d = -3

Therefore, Common difference is -3 and the number of terms are 12.


Exercise 5D Page No: 289

Question 1: 

 

Solution:

Given: (3y – 1), (3y + 5) and (5y+ 1) are in AP

So, (3y + 5) – (3y – 1) = (5y + 1) – (3y + 5)

2 (3y + 5) = (5y + 1) + (3y – 1)

6y + 10 = 8y

8y – 6y = 10

2y = 10

Or y = 5

The value of y is 5.

Question 2:

Solution:

Given: k, (2k – 1) and (2k + 1) are the three successive terms of an AP.

So, (2k – 1) – k = (2k + 1) – (2k – 1)

2 (2k – 1) = 2k + 1 + k

4k – 2 = 3k + 1

4k – 3k = 1 + 2

or k = 3

The value of k is 3.

Question 3:

 

Solution:

Given: 18, a, (b – 3) are in AP

a – 18 = b – 3 – a

a + a – b = -3 + 18

2a – b = 15

Question 4:

Solution:

Given: a, 9, b, 25 are in AP.

So, 9 – a = b – 9 = 25 – b

b – 9 = 25 – b

b + b = 22 + 9 = 34

or b = 17

And,

a – b = a – 9

9 + 9 = a + b

a + b = 18

a + 17 = 18

or a = 1

Answer: a = 18, b= 17

Question 5:

Solution:

Given: (2n – 1), (3n + 2) and (6n – 1) are in AP

So, (3n + 2) – (2n – 1) = (6n – 1) – (3n + 2)

(3n + 2) + (3n + 2) = 6n – 1 + 2n – 1

6n + 4 = 8n – 2

8n – 6n = 4 + 2

Or n = 3

Numbers are:

2 x 3 – 1 = 5

3 x 3 + 2 = 11

6 x 3 – 1 = 17

Answer: (5, 11, 17) are required numbers.

Question 6:

 

Solution:

3-digit natural numbers: 100, 101,…….. 990 and

3-digit natural numbers divisible by 7: 105, 112, 119, 126, …, 994

Here, a = 105, d= 7, l = 994

an = (l) = a + (n – 1) d

994 = 105 + (n – 1) x 7

994 – 105 = (n – 1) 7

n – 1 = 127

n = 128

Answer: There are 128 required numbers.

Question 7:

Solution:

3-digit numbers: 100, 101,…….,999

3-digit numbers divisible by 9 : 108, 117, 126, 135, …, 999

Here, a = 108, d= 9, l = 999

an (l) = a + (n – 1) d

999 = 108 + (n – 1) x 9

(n – 1) x 9 = 999 – 108 = 891

n – 1 = 99

n = 100

Question 8:

Solution:

Sum of first m terms of an AP = 2m² + 3m (given)

Sm = 2m² + 3m

Sum of one term = S1 = 2(1)² + 3 x 1 = 2 + 3 = 5 = first term

Sum of first two terms = S2 = 2(2)² + 3 x 2 = 8 + 6=14

Sum of first three terms = S3 = 2(3)² + 3 x 3 = 18 + 9 = 27

Now,

Second term = a2 = S2 – S1 = 14 – 5 = 9

Question 9: 

 

Solution:

AP is a, 3a, 5a,…..

Here, a = a, d = 2a

Sum = Sn = n/2 [2a + (n-1)d]

= n/2[2a + 2an – 2a]

=an2

Question 10:

Solution:

Given AP is 2, 7, 12, 17, …… 47

Here, a = 2, d = 7 – 2 = 5, l = 47

nth term from the end = l – (n – 1 )d

5th term from the end = 47 – (5 – 1) 5 = 47 – 4 x 5 = 27

Question 11:

Solution:

Given AP is 2, 7, 12, 17,……..

Here, a = 2, d = 7 – 2 = 5

Now,

an = a + (n – 1) d = 2 + (n – 1) 5 = 5n – 3

a30 = 2 + (30 – 1) 5 = 2 + 145 = 147 and

a20 = 2 + (20 – 1) 5 = 2 + 95 = 97

a30 – a20 = 147 – 97 = 50

Question 12:

Solution:

Nth term = an = 3n + 5 (given)

a(n-1) = 3 (n – 1) + 5 = 3n + 2

Common difference = d = an – a(n-1)

= (3n + 5) – (3n + 2)

= 3n + 5 – 3n – 2

= 3

Therefore, common difference is 3.

Question 13:

Solution:

Nth term = an = 7 – 4n

a(n-1) = 7 – 4(n – 1) = 11 – 4n

Common difference = d = an – a(n-1)

= (7 – 4n) – (11 – 4n)

= 7 – 4n – 11 + 4n

= -4

Therefore, common difference is -4.

Question 14:

Solution:

Given AP is √8, √18, √32,……..

Above AP can be written as:

2√2 , 3√2, 4 √2, ……

Here a = 2√2 and d = √2

Next term = 4√2 + √2 = 5√2 = √50

Question 15:

Solution:

Given AP is √2, √8, √18,….

Can be written as:

√2, 2√2, 3√2,…..

First term = √2

Common difference = 2√2 – √2 = √2

Next term = 3√2 + √2 = 4√2 = √32

Question 16:
Solution:

Given AP is 21, 18, 15,….

First term = a = 21

Common difference = d = 18-21 = -3

Last term = l = 0

l = a + (n – 1) d

0 = 21 + (n – 1)(-3)

0 = 21 – 3n + 3

24 – 3n = 0

Or n = 8

Answer: Zero is the 8th term.

Question 17:

Solution:

First n natural numbers: 1, 2, 3, 4, 5, …, n

Here, a = 1, d = 1

Sum = Sn = n/2 [2a + (n-1)d]

= n/2 [2(1) + (n-1)(1)]

= n(n+1)/2

Question 18:

Solution:

First n even natural numbers: 2, 4, 6, 8, 10,….,n

Here, a = 2, d = 4 – 2 = 2

Sum = Sn = n/2 [2a + (n-1)d]

= n/2 [2(2) + (n-1)(2)]

= n(n+1)

Question 19: 

 

Solution:

Given:

First term =a = p and

Common difference = d =q

Now,

a10 = a + (n – 1) d

= p + (10 – 1)q

= (p + 9q)

Question 20:

Solution:

AP terms: 4/5, a, 2 (given)

Then,

a – 4/5 = 2 – a

a = 7/5

Question 21:

 

Solution:

Given, 2p + 1, 13, 5p – 3 are in AP

Then,

13 – (2p + 1) = (5p – 3) – 13

13 – 2p – 1 = 5p – 3 – 13

12 – 2p = 5p – 16

p = 4

The value of p is 4.

Question 22:

Solution:

Given, (2p – 1), 7, 3p are in AP

Then,

7 – (2p – 1) = 3p – 7

7 – 2p + 1 = 3p – 7

5p = 15

p = 3

The value of p is 3.

Question 23:

Solution:

Sum of first p terms = Sp = (ap² + bp)

Sum of one term = S1 = a(1)² + b(1) = a+b = first term

Sum of first two terms = S2 = a(2)² + b x 2 = 4a + 2b

We know that, second term = a2 = S2 – S1

= (4a + 2b) – (a + b)

= 3a + b

Now, d = a2 – a1

= 3a + b – (a+b)

= 2a

Answer: Common difference is 2a.

Question 24:

Solution:

Sum of first n terms = Sn = (3n² + 5n)

Sum of one term = S1 = 3(1)² + 5(1) = 8 = first term

Sum of first two terms = S2 = 3(2)² + 5(2) = 22

We know that, second term = a2 = S2 – S1

=22 – 8

= 14

=> a2 = 14

Now, d = a2 – a1

= 14 – 8 = 6

Answer: Common difference is 6.

Question 25:

Solution:

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

Given:

4th term = a4 = 9

Sum of 6th and 13th terms = a6 + a13 = 40

Now,

a4 = a + (4-1)d

9 = a + 3d

a = 9 – 3d ….(1)

And

a6 + a13 = 40

a + 5d + a + 12d = 40

2a +17d = 40

2(9 – 3d) + 17d = 40 (using (1))

d = 2

From (1): a = 9 – 6 = 3

Required AP = 3,5,7,9,…..

Question 26: 

 

Solution:

Given: a27 – a7 = 84

[a + 26d] – [a + 6d] = 84

20d = 84

d = 4.2

Question 27:

Solution:

Given: 1 + 4 + 7 + 10 + … + x = 287

Here a = 1, d = 3 and Sn = 287

Sum = Sn = n/2 [2a + (n-1)d]

287 = n/2 [2 + (n-1)3]

574 = 3n2 – n

Which is a quadratic equation.

Solve 3n2 – n – 574 = 0

3n2 – 42n + 41n – 574 = 0

3n(n – 14) + 41(n-14) = 0

(3n + 41)(n-14) = 0

Either (3n + 41) = 0 or (n-14) = 0

n = -41/3 or n = 14

Since number of terms cannot be negative, so result is n = 14.

=> Total number of terms in AP are 14.

Which shows, x = a14

or x = a + 13d

or x = 1 + 39

or x = 40

The value of x is 40.


R S Aggarwal Solutions for Chapter 5 Arithmetic Progression

In this chapter students will study important concepts on arithmetic progression as listed below:

  • Arithmetic Progression introduction
  • Arithmetic series
  • Finding the general term of an AP
  • Finding the nth term from the end of an AP
  • Arithmetic Mean
  • Sum of n terms of an AP

Key Features of R S Aggarwal Solutions for Class 10 Maths Chapter 5 Arithmetic Progression

1. R S Aggarwal Solutions for Maths is easy to comprehend and covers all the exercise questions.

2. R S Aggarwal textbook helps to get to know more about various concepts on Arithmetic Progression.

3. Step-by-step problem solving approach helps students to understand easily.

4. Easy for quick revision.

5. Prepared based on the latest CBSE syllabus by subject experts.

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