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

## Chapter 11 - Arithmetic Progressions Ex. 5B

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)

If are consecutive terms of an AP, then

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)

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)

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)

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°

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)

## Chapter 11 - Arithmetic Progressions Ex. 5C

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

Now, 20^{th} 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 20^{th}term is 99

(i)The nth term is given by

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

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

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

Let the required number of terms be n, then

sum of first 15 terms = 0

The term AP is

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

Thus, common difference = 3

The given AP is

Common difference d =

Term next to

The given AP is

First term

Common difference d =

Sum of n terms =

First term 'a' of an AP = 2

The last term l = 29

common difference = 3

First term of an AP, a = 22

Last term = n^{th} term = - 11

Thus, n = 12, d = -3

## Chapter 11 - Arithmetic Progressions MCQ

## Chapter 5 - Arithmetic Progressions Ex. 5A

The given AP is

First term = 6, common difference =

a = 6, d =

The n^{th} term is given by

Hence, 37^{th} term is 69

The given AP is

The first term = 5,

common difference =

a = 5,

The n^{th} term is given by

Hence the 25^{th} term is - 7

(i)First term = -6

(ii)Common difference

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

= (-6 + 15 4) = 54

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

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

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

Suppose there are n terms in given AP, then

Hence, the 18^{th} term of given AP is 88

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 19^{th} term in the given AP is 0

In the given AP, we have

Suppose there are n terms in given AP, we have

Then,

Thus, 14^{th} term in the given AP is 3

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

a = 5, d = 15 - 5 = 10

Thus, the required term is 44^{th}

In the given AP let the first term = a,

And common difference = d

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

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

And n = 8

Hence, the 8^{th} term from the end is 163

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

And n = 6

Now, n^{th} term from the end = [l - (n - 1)d]

Hence, the 6^{th} term from the end is - 25

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

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

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

Hence 25^{th} term is triple its 11^{th} term

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

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

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

n^{th}term = 63 + (n - 1) 2 = 63 + 2n - 2 = 2n + 61

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

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

n^{th} term = 3 + (n - 1) 7 = 3 + 7n - 7 = 7n - 4

The two n^{th} terms are equal

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

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

p^{th} term = a +(p - 1)d = q(given)-----(1)

q^{th} 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)

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

n^{th} term from the beginning = a + (n - 1)d-----(1)

n^{th} term from end= l - (n - 1)d ----(2)

adding (1) and (2),

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

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 n^{th} term be the last term

Hence, there are 17 rows in the flower bed.

## Chapter 5 - Arithmetic Progressions Ex. 5D

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

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

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

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

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

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

Let n^{th} term be zero

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

21 - 3n + 3 = 0

3n = 24

or

Hence, 8^{th} term of given series is 0

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

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

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

a = 2, d = 4 - 2 = 2

First term of AP = a = p

Common difference = d = q

n^{th} term = a + (n 1)d

10^{th} term = p + (10 1)q

= p + 9q

### 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

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