# 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 BYJUâ€™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, â€¦

Here, 15 â€“ 9 = 21 â€“ 15 = 27 â€“ 21 = 6 (which is constant)

Common difference is 6

Or d = 6

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

(ii) 11, 6, 1, -4, â€¦

Here, 6 â€“ 11 = 1 â€“ 6 = -5 -4 â€“ 1 = -5 (which is constant)

d (common difference) = -5

Next term = -4 â€“ 5 = -9

(iii) -1, -5 /6 , -2 /3 , -1 /2 , â€¦â€¦â€¦â€¦

-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) âˆš2, âˆš8, âˆš18, âˆš32, â€¦â€¦..

âˆš8 – âˆš2 = 2âˆš2 – âˆš2 = âˆš2

âˆš32 – âˆš18 = 4âˆš2 – 3âˆš2 = âˆš2

d (common difference) = âˆš2

Next term = âˆš32 + âˆš2 = 4âˆš2 + âˆš2 = 5âˆš2 = âˆš50

(v) âˆš20, âˆš45, âˆš80, âˆš125, â€¦â€¦

âˆš20, âˆš45, âˆš80, âˆš125, â€¦â€¦

âˆš45 – âˆš20 = 3âˆš5 – 2âˆš5 = âˆš5

âˆš125 – âˆš80 = 5âˆš5 – 4âˆš5 = âˆš5

d (common difference) = âˆš5

Next term = âˆš125 + âˆš5 = 5âˆš5 + âˆš5 = 6âˆš5 or âˆš180

Question 2:

Solution:

(i)Given: AP is 9, 13, 17, 21, â€¦â€¦

Here, first term = a = 9

Common difference = d = 13 â€“ 9 = 4

an = a + (n-1)d

a20= 9 + (20-1)4

= 85

(ii) the 35th term of the AP 20, 17, 14, 11, â€¦â€¦

Given: AP is 20, 17, 14, 11, â€¦â€¦

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 âˆš2, âˆš18 , âˆš50, âˆš98, â€¦â€¦â€¦

Given: AP is âˆš2, âˆš18 , âˆš50, âˆš98, â€¦â€¦â€¦

or âˆš2, 3âˆš2 , 5âˆš2, 7âˆš2,….

Here, first term = a = âˆš2

Common difference = d = 3âˆš2 – âˆš2 = 2âˆš2

n = 18

an = a + (n-1)d

a18 = âˆš2 + 34âˆš2

= 35âˆš2

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

Given: AP is 3 /4 , 5 /4 , 7 /4 , 9 /4 , â€¦â€¦.

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, â€¦â€¦

Given: AP is -40, -15, 10, 35, â€¦â€¦

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:

n = 25

an = a + (n-1)d

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

= -7

Question 4:

Solution:

If 2p â€“ 1, 3p + 1, 11 are terms in AP, then

a2 â€“ a1 = a3 â€“ a2 â€¦..(1)

From given:

a1 = 2p â€“ 1

a2 = 3p + 1

a3 = 11

From (1), we get

(3p + 1) â€“ (2p â€“ 1) = 11 â€“ (3p + 1)

3p + 1 â€“ 2p + 1 = 11 â€“ 3p â€“ 1

p + 2 = 10 â€“ 3p

4p = 8

p = 2

For p = 2, these terms are in AP.

Question 5:

Solution:

(i)AP is 5, 11, 17, 23, â€¦â€¦

Here, first term = a = 5

Common difference = d = 11 â€“ 5 = 6

Now,

an = a (n â€“ 1)d

= 5 + (n â€“ 1) 6

= 5 + 6n â€“ 6

= (6n â€“ 1)

(ii) AP is 16, 9, 2, -5, â€¦â€¦

Here, first term = a = 16

Common difference = d = 9 â€“ 16 = -7

an = a + (n â€“ 1)d

= 16 + (n â€“ 1) (-7)

= 16 â€“ 7n + 7

= (23 â€“ 7n)

Question 6:

Solution:

nth term of AP is 4n â€“ 10 (Given)

Putting n = 1, 2, 3, 4, â€¦, we get

At n = 1: 4n â€“ 10 = 4 x 1 â€“ 10 = 4 â€“ 10 = -6

At n = 2: 4n â€“ 10 = 4 x 2 â€“ 10 = 8 â€“ 10 = -2

At n = 3: 4n â€“ 10 = 4 x 3 â€“ 10 = 12 â€“ 10 = 2

At n = 1: 4n â€“ 10 = 4 x 4 â€“ 10 = 16 â€“ 10 = 6

We see that -6, -2, 2, 6,â€¦ are in AP

(i) first term = -6

(ii) Common difference = -2 â€“ (-6) = 4

(iii) 16th term:

Using formula: an = a + (n â€“ 1)d

Here n = 16

a16 = -6 + (16 â€“ 1)4 = 54

Question 7:

Solution:

Given: AP is 6, 10, 14, 18,â€¦, 174

Here, first term = a = 6

Common difference = d= 10 â€“ 6 = 4

To find: the number of terms (n)

Last term = a + (n â€“ 1)d

174 = 6 + (n â€“ 1) 4

174 â€“ 6 = (n â€“ 1) 4

n â€“ 1 = 168 /4 = 42

n = 42 + 1 = 43

There are 43 terms.

Question 8:

Solution:

Given: AP is 41, 38, 35,â€¦, 8

Here, first term = a = 41

Last term = 8

Common difference = d = 38 â€“ 41 = -3

To find: the number of terms (n)

Last term = a + (n â€“ 1)d

8 = 41 + (n â€“ 1)(-3)

8 â€“ 41 = (n â€“ 1)(-3)

n â€“ 1 = 11

n = 11 + 1 = 12

There are 12 terms.

Question 9:

Solution:

Given: AP is 18, 31/2 , 13, â€¦, -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 â€“ 1)d

-47 = 18 + (n â€“ 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, â€¦

Here,

First term = a = 3

Common difference = d = 8 â€“ 3 = 5

nth term of AP is an = a + (n â€“ 1) d

Now,

88 = 3 + (n â€“ 1)(5)

88 â€“ 3 = (n â€“ 1) x 5

n â€“ 1 = 88 /5

or n = 17 + 1 = 18

Therefore: 88 is the 18th term.

Question 11:

Solution:

AP is 72, 68, 64, 60, â€¦..

Let nth term is 0.

Here,

First term = a = 72

Common difference = d = 68 â€“ 72 = -4

an = a + (n â€“ 1)d

0 = 72 + (n â€“ 1)(-4)

-72 = -4(n â€“ 1)

n â€“ 1 = 18

n = 18 + 1 = 19

Therefore: 0 is the 19th term.

Question 12:

Solution:

Here,

First term = a = 5/6

Common difference = d = 1 â€“ 5/6 = 1/6

Now: an = a + (n â€“ 1)d

3 = 5/6 + (n â€“ 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, â€¦â€¦

Let nth term -81

Here, a = 21, d = 18 â€“ 21 = -3

an = a + (n â€“ 1)d

-81 = 21 + (n â€“ 1)(-3)

-81 â€“ 21 = (n â€“ 1)(-3)

-102 = (n â€“ 1)(-3)

n = 34 + 1 = 35

Therefore, -81 is the 35th term

Question 14:

Solution:

Given AP is 3, 8, 13, 18,â€¦

First term = a = 3

Common difference = d = 8 â€“ 3 = 5

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

a20 = a + (n â€“ 1)d

= 3 + (20 â€“ 1) 5

= 3 + 95

= 98

The required term = 98 + 55 = 153

Now, 153 be the nth term, then

an = a + (n â€“ 1)d

153 = 3 + (n â€“ 1) x 5

153 â€“ 3 = 5(n â€“ 1)

150 = 5(n â€“ 1)

n â€“ 1 = 30

n = 31

Required term will be 31st term.

Question 15:

Solution:

AP is 5, 15, 25,â€¦

First term = a = 5

Common difference = d = 15 â€“ 5 = 10

Find 31st term:

a31 = a + (n â€“ 1)d

= 5 + (31 â€“ 1) 10

= 5 + 30 x 10

= 305

Required term = 305 + 130 = 435

Now, say 435 be the nth term, then

an = a + (n â€“ 1)d

435 = 5 + (n â€“ 1)10

435 â€“ 5 = (n â€“ 1)10

n â€“ 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

Question 17:

a16 = 6 + (16 â€“ 1)7 = 6 + 105 = 111

Therefore, midterm of the AP id 111.

Question 18:

Solution:

AP is 10, 7, 4, â€¦..â€¦, (-62)

a = 10,

d = 7 â€“ 10 = -3, and

l = -62

Now, an = l = a + (n â€“ 1)d

-62 = 10 + (n â€“ 1) x (-3)

-62 â€“ 10 = -3(n- 1)

-72 = -3(n â€“ 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 â€“ 1)(-3) = 10 â€“ 36 = -26

Question 19:

Solution:

Given AP is -4 /3 , -1, -2 /3 , â€¦, 13 /3

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,â€¦, 184

a = 7, d = 10 â€“ 7 = 3 and l = 184

nth term from the end = l â€“ (n â€“ 1)d

Now,

8th term from the end be

184 â€“ (8 â€“ 1)3 = 184 â€“ 21 = 163

Question 21:

Solution:

Given: AP is 17, 14, 11, â€¦,(-40)

a = 17, d = 14 â€“ 17 = -3, l = -40

6th term from the end = l â€“ (n â€“ 1)d

= -40 – (6 – 1) (-3)

= -40 â€“ (5 x (-3))

= -40 + 15

= -25

Question 22:

Solution:

Given AP is 3, 7, 11, 15, â€¦

a = 3, d = 7 â€“ 3 = 4

Let 184 be the nth term of the AP

an = a + (n â€“ 1)d

184 = 3 + (n â€“ 1) x 4

184 â€“ 3 = (n â€“ 1) x 4

181 /4 = n â€“ 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,â€¦

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

Let -150 be the nth term of the AP

an = a + (n â€“ 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,â€¦ is negative. Let Tn be the nth term then

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

Question 25:

Solution:

a = 20, d = -3/4

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

an < 0

an = a + (n â€“ 1)d

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 â€“ 1)d

a7 = a + (7 â€“ 1)d

= a + 6d = -4 â€¦â€¦â€¦â€¦(1)

And a13 = a + 12d = -16 â€¦..â€¦..(2)

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

6d = -16 â€“ (-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, â€¦â€¦

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 â€“ 1)d

= a + 3d

Since 4th term of an AP is zero.

a + 3d = 0

or a = -3d â€¦.(1)

Similarly,

a25 = a + 24d = -3d + 24d = 21d â€¦(2)

a11 = a + 10d = -3d + 10d = 7d â€¦.(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 â€“ 1 )d

a + (15 â€“ 1)d = -5d + 14d = 9d

and a33 = a + (n â€“ 1 )d = a + (33 â€“ 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 â€“ 1)d

a4 = a + (4 â€“ 1)d = a + 3d

a + 3d = 11 â€¦â€¦â€¦(1)

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

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

2a + 10d = 34

a + 5d = 17 â€¦â€¦..(2)

Subtracting (1) from (2), we get

2d = 17 â€“ 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 â€“ 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 â€“ 2), (4k â€“ 6) and (k + 2) are three consecutive terms of an AP.

So, (4k â€“ 6) â€“ (3k â€“ 2) = (k + 2) â€“ (4k â€“ 6)

2(4k â€“ 6) = (k + 2) + (3k â€“ 2)

8k â€“ 12 = 4k + 0

8k â€“ 4k = 0 + 12

or k = 3

Question 2:

Solution:

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

So, d = (4x â€“ 1) â€“ (5x + 2) = (x + 2) â€“ (4x â€“ 1)

2(4x â€“ 1) = (x + 2) + (5x + 2)

8x â€“ 2 = 6x + 2 + 2

8x â€“ 2 = 6x + 4

8x â€“ 6x = 4 + 2

or x = 3

The value of x is 3.

Question 3:

Solution:

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

So, (3y + 5) â€“ (3y â€“ 1) â€“ (5y + 1) â€“ (3y + 5)

2(3y + 5) = 5y + 1 + 3y â€“ 1

6y + 10 = 8y

8y â€“ 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 â€“ (x + 2) = (2x + 3) â€“ 2x

2x â€“ x â€“ 2 = 2x + 3 â€“ 2x

x â€“ 2 = 3

x = 2 + 3 = 5

The value of x is 5.

Question 5:

Solution:

Assume that (a â€“ b)Â², (aÂ² + bÂ²) and (a + b)Â² are in AP.

So, (aÂ² + bÂ²) â€“ (a â€“ b)Â² = (a + b)Â² â€“ (aÂ² + bÂ²)

(aÂ² + bÂ²) â€“ (aÂ² + bÂ² â€“ 2ab) = aÂ² + bÂ² + 2ab â€“ aÂ² â€“ bÂ²

2ab = 2ab

Which is true.

Hence given terms are in AP.

Question 6:

Solution:

Let a â€“ d, a, a + d are three numbers in AP.

So, their sum = a â€“ d + a + a + d = 15

3a = 15

or a = 5

Again,

Their Product = (a â€“ d) x a x (a + d) = 80

a(aÂ² â€“ dÂ²) = 80

5(5Â² â€“ dÂ²) = 80

25 â€“ dÂ² = 16

dÂ² = 25 â€“ 16 = 9 = (Â±3)Â²

or d = Â±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 â€“ d, a, a + d are three numbers in AP.

their sum = a â€“ d + a + a + d = 3

3a = 3

or a = 1

Again,

Their Product = (a â€“ d) x a x (a + d) = -35

a(aÂ² â€“ dÂ²) = -35

(1Â² â€“ dÂ²) = -35

or d = Â±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 â€“ d, a, a + d are three numbers in AP.

their sum = a â€“ d + a + a + d = 24

3a = 24

or a = 8

Again,

Their Product = (a â€“ d) x a x (a + d) = 440

a(aÂ² â€“ dÂ²) = 440

8(8Â² â€“ dÂ²) = 440

or d = Â±3

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

Question 9:

Solution:

Let a â€“ d, a, a + d are three numbers in AP.

their sum = a â€“ d + a + a + d = 21

3a = 21

or a = 7

Again,

Sum of squares = (a â€“ d)^2 + a^2 +(a + d)^2 = 165

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

Question 10:

Solution:

Sum of angles of a quadrilateral = 360Â°

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 = 360Â°

4a + 60 = 360

4a = 300

or a = 75Â°

Other angles:

a + 10 = 75 + 10 = 85

a + 20 = 75 + 20 = 95

a + 30 = 75 + 30 = 105

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

Question 11:

Solution:

Let a â€“ 3d, a â€“ d, a + d, a + 3d are the four numbers in AP.

Their sum = a â€“ 3d + a â€“ d + a + d + a + 3d = 28

Sum of their square = (a â€“ 3d)^2 + (a â€“ d)^2 + (a + d)^2 + (a + 3d)^2 = 216

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

Question 12:

Solution:

Let a â€“ 3d, a â€“ d, a + d, a + 3d are the four numbers in AP.

Therefore:

d = Â±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 â€“ d, a, a + d are the three terms in AP

So,

Sum = a â€“ d + a + a + d = 48

3a = 48

Or a = 16

And,

a(a â€“ d) = 4 (a + d) + 12

16 (16 â€“ d) = 4(16 + d) + 12

256 â€“ 16d = 64 + 4d + 12 = 4d + 76

256 â€“ 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:

â€¦â€¦(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 Ã— 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 Ã— (-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) Ã— [(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) Ã— [1.2 + (99 Ã— 1.1) ]

= 50 Ã— 110.1

= 5505

Sum of 100 terms of this AP is 5505.

Question 2:

Solution:

(i) 7 + 10 1/2 + 14 + â€¦ + 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 + â€¦ + 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) + â€¦ + (-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 Ã— [(-10) + (75)(-3)]

= – 8930

Sum of 76 terms of this AP is -8930.

(iv) 5 + (-41) + 9 + (-39) + 13 + (-37) + 17 + â€¦ + (-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:Â

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 = 3nÂ² + 6n

Question 5:

Solution:

Sn = 3nÂ² â€“ n

S1 = 3(1)Â² â€“ 1 = 3 â€“ 1 = 2

S2 = 3(2)Â² â€“ 2 = 12 â€“ 2 = 10

a1 = 2

a2 = 10 â€“ 2 = 8

(i) an = a + (n â€“ 1) d

= 2 + (n â€“ 1) x 6

= 2 + 6n â€“ 6

= 6n â€“ 4

(ii) First term = 2

(iii) Common difference = 8 â€“ 2 = 6

Question 6:Â

Solution:

(ii)

Question 7:Â

Solution:

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

mth term = 1/n

so,

am = a + (m â€“ 1) d = 1/n ……….(1)

nth term = 1/m

an = a + (n â€“ 1) d = 1/m ……….(2)

Subtract (2) from (1)

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,â€¦

a = 21,

d = 18 â€“ 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:Â

Solution:

AP is 9, 17, 25,â€¦

a = 9, d = 17 â€“ 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 canâ€™t be negative and fraction, so

n = 12

Number of terms = 12 terms.

Question 10:Â

Solution:

AP is 63, 60, 57, 54,â€¦

Here, a = 63, d = 60 â€“ 63 = -3 and sum = Sn = 693

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)

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

= 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, â€¦, 49

Here, a = 1, d = 3 â€“ 1 = 2, l = 49

Sum of odd numbers:

= n/2(a + l)

= 25/2 (l + 49)

= 25/2(50)

= 625

Question 13:Â

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

Sum of all 29 terms is 8729.

Question 14:Â

Solution:

First forty positive integers which are divisible by 6 are

6, 12, 18, 24,…… to 40 terms

Here, a = 6, d = 12 â€“ 6 = 6, and n = 40.

Sum of 40 terms:

Question 15:

Solution:

First 15 multiples of 8 are as given below:

8, 16, 24, 32,….. to 15 terms

Here, a = 8, d = 16 â€“ 8 = 8, and n = 15

Sum of 15 terms:

Question 16:Â

Solution:

Multiples of 9 lying between 300 and 700 = 306, 315, 324, 333, â€¦, 693

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

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

There are 44 terms.

Find the sum of 44 terms:

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 â€“ 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:Â

Solution:

Even natural numbers: 2, 4, 6, 8, 10, â€¦

Even natural numbers divisible by 5: 10, 20, 30, 40, â€¦ to 100 terms

Here, a = 10, d = 20 â€“ 10 = 10, and n = 100

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

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

= 50500

Question 19:

(4 â€“ 1/n) + (4-2/n) + (4â€“ 3/n) + â€¦.

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:Â

Solution:

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

Which implies: a = 1 and d = 5

Therefore, the AP is 1, 6, 11, 16,â€¦â€¦

Question 21:Â

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:Â

Solution:

Given: (3y â€“ 1), (3y + 5) and (5y+ 1) are in AP

So, (3y + 5) â€“ (3y â€“ 1) = (5y + 1) â€“ (3y + 5)

2 (3y + 5) = (5y + 1) + (3y â€“ 1)

6y + 10 = 8y

8y â€“ 6y = 10

2y = 10

Or y = 5

The value of y is 5.

Question 2:

Solution:

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

So, (2k â€“ 1) â€“ k = (2k + 1) â€“ (2k â€“ 1)

2 (2k â€“ 1) = 2k + 1 + k

4k â€“ 2 = 3k + 1

4k â€“ 3k = 1 + 2

or k = 3

The value of k is 3.

Question 3:

Solution:

Given: 18, a, (b â€“ 3) are in AP

a â€“ 18 = b â€“ 3 â€“ a

a + a â€“ b = -3 + 18

2a â€“ b = 15

Question 4:

Solution:

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

So, 9 â€“ a = b â€“ 9 = 25 â€“ b

b â€“ 9 = 25 â€“ b

b + b = 22 + 9 = 34

or b = 17

And,

a â€“ b = a â€“ 9

9 + 9 = a + b

a + b = 18

a + 17 = 18

or a = 1

Answer: a = 18, b= 17

Question 5:

Solution:

Given: (2n â€“ 1), (3n + 2) and (6n â€“ 1) are in AP

So, (3n + 2) â€“ (2n â€“ 1) = (6n â€“ 1) â€“ (3n + 2)

(3n + 2) + (3n + 2) = 6n â€“ 1 + 2n â€“ 1

6n + 4 = 8n â€“ 2

8n â€“ 6n = 4 + 2

Or n = 3

Numbers are:

2 x 3 â€“ 1 = 5

3 x 3 + 2 = 11

6 x 3 â€“ 1 = 17

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

Question 6:

Solution:

3-digit natural numbers: 100, 101,â€¦â€¦.. 990 and

3-digit natural numbers divisible by 7: 105, 112, 119, 126, â€¦, 994

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

an = (l) = a + (n â€“ 1) d

994 = 105 + (n â€“ 1) x 7

994 â€“ 105 = (n â€“ 1) 7

n â€“ 1 = 127

n = 128

Answer: There are 128 required numbers.

Question 7:

Solution:

3-digit numbers: 100, 101,â€¦â€¦.,999

3-digit numbers divisible by 9 : 108, 117, 126, 135, â€¦, 999

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

an (l) = a + (n â€“ 1) d

999 = 108 + (n â€“ 1) x 9

(n â€“ 1) x 9 = 999 â€“ 108 = 891

n â€“ 1 = 99

n = 100

Question 8:

Solution:

Sum of first m terms of an AP = 2mÂ² + 3m (given)

Sm = 2mÂ² + 3m

Sum of one term = S1 = 2(1)Â² + 3 x 1 = 2 + 3 = 5 = first term

Sum of first two terms = S2 = 2(2)Â² + 3 x 2 = 8 + 6=14

Sum of first three terms = S3 = 2(3)Â² + 3 x 3 = 18 + 9 = 27

Now,

Second term = a2 = S2 â€“ S1 = 14 â€“ 5 = 9

Question 9:Â

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, â€¦â€¦ 47

Here, a = 2, d = 7 â€“ 2 = 5, l = 47

nth term from the end = l â€“ (n â€“ 1 )d

5th term from the end = 47 â€“ (5 â€“ 1) 5 = 47 â€“ 4 x 5 = 27

Question 11:

Solution:

Given AP is 2, 7, 12, 17,â€¦â€¦..

Here, a = 2, d = 7 â€“ 2 = 5

Now,

an = a + (n â€“ 1) d = 2 + (n â€“ 1) 5 = 5n â€“ 3

a30 = 2 + (30 â€“ 1) 5 = 2 + 145 = 147 and

a20 = 2 + (20 â€“ 1) 5 = 2 + 95 = 97

a30 â€“ a20 = 147 â€“ 97 = 50

Question 12:

Solution:

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

a(n-1) = 3 (n â€“ 1) + 5 = 3n + 2

Common difference = d = an â€“ a(n-1)

= (3n + 5) â€“ (3n + 2)

= 3n + 5 â€“ 3n â€“ 2

= 3

Therefore, common difference is 3.

Question 13:

Solution:

Nth term = an = 7 â€“ 4n

a(n-1) = 7 â€“ 4(n â€“ 1) = 11 â€“ 4n

Common difference = d = an â€“ a(n-1)

= (7 â€“ 4n) â€“ (11 â€“ 4n)

= 7 â€“ 4n â€“ 11 + 4n

= -4

Therefore, common difference is -4.

Question 14:

Solution:

Given AP is âˆš8, âˆš18, âˆš32,……..

Above AP can be written as:

2âˆš2 , 3âˆš2, 4 âˆš2, ……

Here a = 2âˆš2 and d = âˆš2

Next term = 4âˆš2 + âˆš2 = 5âˆš2 = âˆš50

Question 15:

Solution:

Given AP is âˆš2, âˆš8, âˆš18,….

Can be written as:

âˆš2, 2âˆš2, 3âˆš2,…..

First term = âˆš2

Common difference = 2âˆš2 – âˆš2 = âˆš2

Next term = 3âˆš2 + âˆš2 = 4âˆš2 = âˆš32

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 â€“ 1) d

0 = 21 + (n â€“ 1)(-3)

0 = 21 â€“ 3n + 3

24 â€“ 3n = 0

Or n = 8

Answer: Zero is the 8th term.

Question 17:

Solution:

First n natural numbers: 1, 2, 3, 4, 5, â€¦, n

Here, a = 1, d = 1

Sum = SnÂ = 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 â€“ 2 = 2

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

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

= n(n+1)

Question 19:Â

Solution:

Given:

First term =a = p and

Common difference = d =q

Now,

a10 = a + (n â€“ 1) d

= p + (10 â€“ 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 â€“ 3 are in AP

Then,

13 â€“ (2p + 1) = (5p â€“ 3) â€“ 13

13 â€“ 2p â€“ 1 = 5p â€“ 3 â€“ 13

12 â€“ 2p = 5p â€“ 16

p = 4

The value of p is 4.

Question 22:

Solution:

Given, (2p â€“ 1), 7, 3p are in AP

Then,

7 â€“ (2p â€“ 1) = 3p â€“ 7

7 â€“ 2p + 1 = 3p â€“ 7

5p = 15

p = 3

The value of p is 3.

Question 23:

Solution:

Sum of first p terms = Sp = (apÂ² + bp)

Sum of one term = S1 = a(1)Â² + b(1) = a+b = first term

Sum of first two terms = S2 = a(2)Â² + 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

Question 24:

Solution:

Sum of first n terms = Sn = (3nÂ² + 5n)

Sum of one term = S1 = 3(1)Â² + 5(1) = 8 = first term

Sum of first two terms = S2 = 3(2)Â² + 5(2) = 22

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

=22 – 8

= 14

=> a2 = 14

Now, d = a2 – a1

= 14 – 8 = 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:Â

Solution:

Given: a27 â€“ a7 = 84

[a + 26d] – [a + 6d] = 84

20d = 84

d = 4.2

Question 27:

Solution:

Given: 1 + 4 + 7 + 10 + â€¦ + 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

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.