Competitive Exams

IIT JEE, NEET, ESE, GATE, AE/JE, Olympiad

Only IAS

UPSC, State PSC

School Preparation

Foundation (Class 6-12), Commerce, Arts, CuriousJr (1st - 8th), Science, International Boards

Govt Exam

Judiciary, SSC, Defence, Teaching, JAIIB & CAIIB, BIHAR EXAMS WALLAH, UP Exams, Railway, Nursing Exams, Banking, WB Exams

UG & PG Entrance Exams

MBA, IPMAT, IIT JAM, LAW, CUET UG, UGC NET, GMAT, Design & Architecture, Pharma, CSIR NET, CUET PG, NEET PG

FINANCE

CA, CS, Finance Courses, ACCA

Others

English Wallah, Online Degrees

English Proficiency Test

IELTS, TOEFL

Agriculture

RS Aggarwal Solutions for Class 10 Maths Chapter 11 Exercise 11.3 Arithmetic Progressions

In this article we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 11 Exercise 11.3 prepared by our experts to help students to prepare better for their examinations.

Ananya Gupta18 Jul, 2024

RS Aggarwal Solutions for Class 10 Maths Chapter 11 Exercise 11.3 Arithmetic Progressions

RS Aggarwal Solutions for Class 10 Maths Chapter 11 Exercise 11.3: In Chapter 11 of RS Aggarwal's Class 10 Maths book, which covers Arithmetic Progressions (AP), Exercise 11.3 focuses on determining the number of terms in an AP and finding specific terms based on given information. The exercise involves calculations where students use formulas such as the nth term formula and the formula for the last term of an AP.

These exercises are designed to strengthen understanding of AP concepts and enhance problem-solving skills through practical applications of AP formulas.

RS Aggarwal Solutions for Class 10 Maths Chapter 11 Arithmetic Progressions Exercise 11.3 Overview

The solutions for Chapter 11 Exercise 11.3 of RS Aggarwal's Class 10 Maths have been prepared by subject experts from Physics Wallah. This exercise focuses on Arithmetic Progressions (AP), where students learn to find terms in a sequence and solve problems using basic AP formulas. Each solution is designed to be easy to follow, giving clear steps to help students understand and practice these mathematical concepts effectively.

RS Aggarwal Solutions for Class 10 Maths Chapter 11 Exercise 11.3 PDF

You can access the PDF link for RS Aggarwal Solutions for Class 10 Maths, Chapter 11 Exercise 11.3 below. This PDF provides detailed solutions prepared by subject experts, helping students understand Arithmetic Progressions (AP) and solve problems effectively.

RS Aggarwal Solutions for Class 10 Maths Chapter 11 Exercise 11.3 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 11 Exercise 11.3

Here we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 11 Arithmetic Progressions Exercise 11.3 for the ease of students so that they can prepare better for their exams.

Q. The sum of the first n terms of an AP is $(3 n^{2} + 6 n) .$ Find the nth term and the 15th terms of this AP.

Solution:

Here, first term = 9 Common difference = 7 - 9 = - 2 Sum of first n terms of an AP is Sn = n 2 [2a + (n - 1)d] ∴ S14 = 14 2 [2(9) + (14 - 1)(-2)] = (7)(18 - 26) = (7) × (-8) = - 56 Thus, sum of 14 terms of this AP is - 56.

Q. (i) the sum for the first n terms of an AP is $(\frac{5 n^{2}}{2} + \frac{3 n}{2})$ . Find hte nth term and the 20th term of this AP.

(ii) The sum of the first n terms of an AP is $(\frac{3 n^{2}}{2} + \frac{5 n}{2})$ . Find its nth term and the 25th term.

Solution:

(i) Sum of n terms, $S_{n} = \frac{5 n^{2}}{2} + \frac{3 n}{2}$

S_{1} = \frac{5 (1)^{2}}{2} + \frac{3 (1)}{2} = \frac{5}{2} + \frac{3}{2} = \frac{8}{2} = 4 = a

S_{1} = \frac{5 (2)^{2}}{2} + \frac{3 (2)}{2} = \frac{20}{2} + \frac{6}{2} = 10 + 3 = 13

a_{2} = S_{2} - S_{1} = 13 - 4 = 9

d = a_{2} - a_{1} = 9 - 4 = 5

a_{n} = a + (n - 1) d = 4 + (n - 1) 5 = 4 + 5 n - 5 = 5 n - 1

a_{20} = 5 n - 1 = 5 \times 20 - 1 = 100 - 1 = 99

(ii) Sum of n terms, $S_{n} = \frac{3 n^{2}}{2} + \frac{5 n}{2}$

S_{1} = \frac{3 (1)^{2}}{2} + \frac{5 (1)}{2} = \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4 = a

S_{1} = \frac{3 (2)^{2}}{2} + \frac{3 (2)}{2} = \frac{12}{2} + \frac{10}{2} = 6 + 5 = 11

a_{2} = S_{2} - S_{1} = 11 - 4 = 7

d = a_{2} - a_{1} = 7 - 4 = 3

a_{n} = a + (n - 1) d = 4 + (n - 1) 3 = 4 + 3 n - 3 = 3 n + 1

a_{25} = 3 n + 1 = 3 \times 25 + 1 = 75 + 1 = 76

Q. How many terms of the AP: 21, 18, 15, ... must be added to get the sum 0?

Solution:

Q. How many terms of the AP 9, 17, 25, ... must be taken so that their sum is 636?

Solution:

A.P = 9,17,25 a=9 d=17-9 =8 Sum of n terms,

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

636 = \frac{n}{2} [2 (9) + (n - 1) 8]

636 = \frac{n}{2} [18 + 8 n - 8]

636 = \frac{n}{2} [10 + 8 n]

1272 = 10 n + 8 n^{2}

8 n^{2} + 10 n - 1272 = 0

2 [4 n^{2} + 5 n - 636] = 0

4 n^{2} + 5 n - 636 = 0

4 n^{2} + 53 n - 48 n - 636 = 0

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

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

n - 12 = 0

n = 12

12 terms must be given to sum of 636

Q. How many terms of the AP 63, 60, 57, 54, ... must be taken so that their sum is 693. Explain the double answer.

Solution:

Given, AP 63, 60, 57 where, a = 63 and the difference (d) = 60 - 63 = -3 also given that

S_{n}

= 693 ∴ to find a, we know

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

By substituting the values of a, d and

S_{n}

we get; 693 =

\frac{n}{2}

[ 2 ×63 + (n - 1) - 3 ] 693 =

\frac{n}{2}

[ 126 - 3n + 3 ] 693 =

\frac{n}{2}

[ 129 - 3n ]

693 = \frac{129 n}{2} - \frac{3 n^{2}}{2}

693 × 2 = 129n -

3 n^{2}

1386 = 129n -

3 n^{2}

1386 - 129n +

3 n^{2}

= 0 By dividing the whole equation by 3 we get,

\frac{1386}{3} - \frac{129 n}{3} + \frac{3 n^{2}}{3} = \frac{0}{3}

462 - 43n +

n^{2}

= 0 ie;

n^{2}

- 43n + 462 = 0 using factorisation method :-- sum = - 43 and product = 462 ∴ the numbers are -21 and -22 So by splitting the middle term we get; (

n^{2}

- 21n ) ( - 22n + 462 ) = 0 n ( n - 21 ) - 22 ( n - 21 ) = 0 ( n - 21 ) ( n - 22 ) = 0 ∴ n = 21 and n = 22 ie; We get the sum of the given AP as 693 when we take first 21 terms of it or 22 terms of the same AP. Verification of the Answer First take n as 21, the

S_{21} = \frac{21}{2} (a + a_{21})

a_{21}

= a + ( 21 - 1 ) d = 63 + [ 20 × ( - 3 ) ] = 63 - 60

a_{21} = 3

∴

S_{21} = \frac{21}{2}

[ 63 + 3 ] =

\frac{21}{2}

× 66

S_{21}

= 693 So the condition is satisfied for when n = 21 Now check for when n = 22

S_{22}

\frac{22}{2}

( a +

a_{22}

)

a_{22}

= a + ( 22 - 1 ) d = 63 + [ 21 × ( -3 ) ] = 63 - 63

a_{22}

= 0 We know

S_{22}

S_{21}

a_{22}

= 693 + 0 = 693 ∴ the condition is satisfied in both the cases so n = 21 or n = 22 Q. Find the sum of all odd numbers between 0 and 50.

Solution:

1st odd no.=1 last odd no.=49 a=1, an or l=49, common diff. d= 2,

a_{n} = a + (n - 1) d

=1+(n-1)2=49 =1+2n-2=49 =2n-1=49 =2n=49=1 =2n=50 n=25

S_{n} = \frac{n}{2} [a + L]

= \frac{25}{2} (1 + 49)

= \frac{25}{2} \times 50

= 25 \times 25 = 625

therefore sum of all odd integers between 0 to 50 is 625

Q. Find the sum of all natural numbers between 200 and 400 which are divisible by 7.

Solution:

The numbers lying between 200 and 400, which are divisible by 7, are 203, 210, 217, … 399 ∴First term, a = 203 Last term, l = 399 Common difference, d = 7 Let the number of terms of the A.P. be n. ∴

a_{n} = a + (n - 1) d

= 399 ⇒ 399 = 203 + (n –1) 7 ⇒ 7 (n –1) = 196 ⇒ n –1 = 28 ⇒ n = 29

S_{n} = \frac{n}{2} [a + L]

S_{29} = \frac{29}{2} [203 + 399]

= \frac{29}{2} \times 602 = 8729

Thus, the required sum is 8729.

Q. Find the sum first forty positive integers divisible by 6.

Solution:

The positive integers divisible by 6 are 6,12,18,....... up to 40 terms the given series is in arthimetic progression with first term a=6 and common difference d=6 the sum of n terms of an A.p is Sum of n terms,

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

= \frac{40}{2} \times 2 (6) + (39) 6

= 20 12 + 234 = 20 \times 246 = 4920

Q. Find the sum of the first 15 multiples of 8.

Solution:

The first 8 multiples of 8 are 8, 16, 24, 32, 40, 48, 56,64.......120 These are in an A.P., having first term as 8 and common difference as 8. Therefore, a = 8 d = 8

S_{15}

= ?

S_{n}

\frac{n}{2}

[2a + (n - 1)d]

\Rightarrow

S_{15}

\frac{15}{2}

[2(8) + (15 - 1)8]

\Rightarrow

S_{15}

= 7.5[6 + (14) (8)]

\Rightarrow

S_{15}

= 7.5[16 + 112]

\Rightarrow

S_{15}

= 7.5(128)

\Rightarrow

S_{15}

= 960

Q. Find the sum of all three - digit natural numbers which are divisible by 13.

Solution:

As per question; AP: 104 , 117 , 130 ........... 988 a = 104 ,

a_{n}

= 988

\Rightarrow

988 = 104 + (n-1) 13

\Rightarrow

988 - 104 = (n-1)13

\Rightarrow

\frac{884}{13}

= n-1

\Rightarrow

n= 69

S_{n}

\frac{n}{2}

(

a + a_{n}

)

\Rightarrow

S_{n}

\frac{n}{2}

( 104 + 988)

\Rightarrow

S_{n}

\frac{69}{2}

( 1092 )

\Rightarrow

S_{n}

= 37674

Q. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Solution:

Let a and d are the first term and the common difference for an AP respectively. Number of terms of AP = n Last term = nth term = l =

a_{n}

a = 17 , d = 9 , l = 350 a + ( n - 1 ) d = 350

\Rightarrow

17 + ( n - 1 ) 9 = 350

\Rightarrow

( n - 1 ) 9 = 350 - 17

\Rightarrow

( n - 1 ) 9 = 333

\Rightarrow

n - 1 =

\frac{333}{9}

\Rightarrow

n - 1 = 37

\Rightarrow

n = 37 + 1

\Rightarrow

n = 38 Therefore, number of terms in given AP = n = 38 Sum of n terms of AP =

S_{n}

S_{n}

( a + l ) Here, n= 38

\Rightarrow

S_{38}

= 19[ 17 + 350 ]

\Rightarrow

S_{38}

= 19 × 367

\Rightarrow

S_{38}

= 6973

Q. In an AP, the first term is 22, nth term is -11 and sum of first n terms is 66. Find n and hence find the common difference.

Solution:

Q. In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees that each section of each class will plant will be double of the class in which they are studying. If there are $1$ to $12$ classes in the school and each class has two sections, find how many trees were planted by students. Which value is shown in the question?

Solution:

Given: The number of trees that each section of each class will plant will be double of the class in which they are studying. Number of classes are

12

(

1

12

), and each class has two sections.

\Rightarrow

Number of trees planted by the students of each section of class

1 = 1 \times 2 = 2

There are two sections of class

1

. Therefore, Number of trees planted by the students of class

1 = 2 \times 2 = 4

Number of trees planted by the students of each section of class

2 = 2 \times 2 = 4

.---(1) There are two sections of class

2

. Therefore, Number of trees planted by the students of class

2 = 2 \times 4 = 8

---(2) Similarly, Number of trees planted by the students of class

3 = 2 \times 6 = 12

---(3) So, the number of trees planted by the students of different classes are

4, 8, 12, . . .

Therefore, Total number of trees planted by the students

= 4 + 8 + 12 + . . . .

up to

12

terms Lets evaluate,

4 + 8 + 12 + . . . .

up to

12

terms. Here,

8 - 4 = 12 - 8 = . . . . = 4

Since, the difference of every consecutive terms is constant. So, this series is an arithmetic series. Here,

a = 4, d = 8 - 4 = 4

and

n = 12

Since, the sum of first

n

- terms of an AP is,,

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

\Rightarrow S_{12} = \frac{12}{2} [2 \times 4 + (12 - 1) \times 4]

\Rightarrow S_{12} = 6 [8 + (11 \times 4)]

\Rightarrow S_{12} = 6 \times [8 + 44]

\Rightarrow S_{12} = 6 \times 52

\Rightarrow S_{12} = 312

Therefore, the total number of trees planted by the students are

312

Q. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are 10 Potatoes in the line. A competitor starts from the bucket, picks up the nearest potato runs back with it, drops it in the bucket, runs back to pick up the next potato runs to the bucket to drop it in, and the continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

Solution:

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 n = 10

S_{10} = ? S_{10} = \frac{12}{2} [2 (10) + (n - 1) (6)] = \frac{12}{2} [2 (10) + (10 - 1) (6)] = 5 [20 + 54] = 5 (74) = 370

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

Q. There are 25 trees at equal distances of 5 m in a line with a water tank, the distance of the water tank from the nearest tree being 10 m. A gardener waters all the trees separately, starting from the water tank and returning back to the water tank after watering each tree to get water for the next, Find the total distance covered by the gardener in order to water all the trees.

Solution:

Distance covered by the gardener to water the first tree and return to the water tank = 10m + 10m = 20 m Distance covered by the gardener to water the second tree and return to the water tank = 15m + 15 m = 30 m Distance covered by the gardener to water the third tree and return to the water tank = 20m + 20m = 40 m Therefore, total distance covered by the gardener to water all the trees = 20m + 30m + 40m + ...... Up to 25 terms This is an arithmetic series. Here a = 20, d = 30 – 20 = 10 and n = 25 Using the formula,

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

we get

S_{25} = \frac{25}{2} [2 \times 20 + (25 - 1) \times 10]

= \frac{25}{2} \times (40 + 240)

= \frac{25}{2} \times 280 = 3500

Hence, the total distance covered by the gardener to water all the trees is 3500 m.

Q. The 24th term of an AP is twice is 10th term. Show that its 72nd term is 4 times its 15 the term.

Solution:

Let first term = a and common difference = d we know that

a_{n} = a + (n - 1) d

Given that,

a_{24} = 2 \times a_{10}

\Rightarrow a + (24 - 1) d = 2 \times (a + (10 - 1) d)

\Rightarrow a + 23 d = 2 \times (a + 9 d)

\Rightarrow a + 23 d = 2 a + 18 d)

\Rightarrow a = 5 d . . . . (i)

Now 15th term,

a_{15} = a + (15 - 1) d = a + 14 d [From (i)] = 5 d + 14 d = 19 d

Similarly 72nd term,

a_{72} = a + (72 - 1) d = a + 71 d = 5 d + 71 d = 76 d

Now,

\frac{a_{72}}{a_{15}} = \frac{76 d}{19 d} = 4

∴ a_{72} = 4 \times a_{15}

Hence,

72^{t h}

term is

4

times to its

15^{t h}

term.

Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 11 Exercise 11.3

Clear Understanding: These solutions provide step-by-step explanations that clarify how to solve problems involving AP, helping students grasp concepts effectively.

Practice: By solving problems from Exercise 11.3 students can practice applying AP formulas and strengthen their problem-solving skills.

Comprehensive Coverage: The solutions cover various types of problems ensuring students are well-prepared for exams and assessments.

Self-assessment: Students can use these solutions to self-assess their progress and identify areas where they may need further practice or clarification.

RS Aggarwal Solutions for Class 10 Maths Chapter 11 Exercise 11.3 FAQs

What does Exercise 11.3 of RS Aggarwal's Class 10 Maths cover?

Exercise 11.3 focuses on determining the number of terms in an Arithmetic Progression (AP) and finding specific terms based on given information such as the first term, common difference, and last term.

How can Exercise 11.3 help in understanding AP?

By practicing Exercise 11.3, students can strengthen their understanding of AP concepts, including how to calculate the number of terms and find specific terms using AP formulas.

Why is it important to practice Exercise 11.3?

Practicing Exercise 11.3 helps students develop confidence in solving AP problems, improves their mathematical reasoning skills, and prepares them thoroughly for exams.

What is an Arithmetic Progression (AP)?

An Arithmetic Progression is a sequence of numbers where the difference between consecutive terms is constant.

🔥 Trending Blogs

NCERT Solutions Federalism Class 10 Questions and Answers PDF

CBSE Class 10 Compartment Result 2025 Anytime @results.cbse.nic.in, Diect Link Here

NCERT Solutions for Class 10 Hindi Kritika Chapter 1 Mata ka Aanchal PDF

NCERT Solutions for Class 10 Geography Chapter 2 Forest Wildlife Resources

NCERT Solutions of Class 10 English First Flight Chapter 4 Question Answer

Talk to a counsellorHave doubts? Our support team will be happy to assist you!

7019243492

07019243492

Check out these Related Articles

CLASS 10 EXAM

NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

Read Full Story

22 Jul 2025, 12:12 pm

CLASS 10 EXAM

NCERT Solutions for Class 10 English First Flight Chapter 2 PDF

Read Full Story

21 Jul 2025, 01:44 pm

CLASS 10 EXAM

NCERT Solutions for Class 10 English First Flight Chapter 3 Question Answer

Read Full Story

19 Jul 2025, 11:13 am

CLASS 10 EXAM

CBSE Class 10 Compartment Admit Card 2025, Check Expected Date

Read Full Story

8 Jul 2025, 04:43 pm

CLASS 10 EXAM

NCERT Solutions for Class 10 Science Chapter 7 Control and Coordination

Read Full Story

5 Jul 2025, 02:36 pm

CLASS 10 EXAM

CBSE Class 10 Compartment Exam Date 2025 OUT Download PDF

Read Full Story

28 Jun 2025, 11:20 am

CLASS 10 EXAM

NCERT Solutions for Class 10 Science Chapter 6 Life Processes

Read Full Story

27 Jun 2025, 08:08 am

CLASS 10 EXAM

CBSE Class 10 Board Exams Conducted Twice a Year from 2026 February Mandatory, May Optional

Read Full Story

26 Jun 2025, 12:38 pm

CLASS 10 EXAM

CBSE Class 10th Toppers List 2025, Rank, Marks, Region Wise Stats other Details

Read Full Story

25 Jun 2025, 01:30 pm

Join 15 Million students on the app today!

Live & recorded classes available at ease

Dashboard for progress tracking

Millions of practice questions at your fingertips

Free Learning Resources

PW Books

Notes (Class 10-12)

Class 10 Math's Notes

Class 10 Chemistry Notes

Class 10 Physics Notes

lass 10 Biology Notes

Aptitude & Reasoning

Class 10 Geography

Physics Class 11 Notes

Class 11 Chemistry Notes

Maths Notes Class 11

Zoology Class 11

Class 11 Botany Notes

Physics Class 12 notes

Chemistry Class 12

Maths Notes Class 12

Zoology class 12

Class 12th Botany Notes

PW Study Materials

Arjuna JEE Study Material

JEE 5 PYQs

JEE Mind Maps

Yakeen Study Material

NCERT Punch

Arjuna Neet Study Material

NEET PYQs

CBSE Sample Paper

CBSE Question Bank

Udaan For Class 10th Study Material

Notes (Class 6-9)

Class-6 Theory & Notes

Math's Notes for class 7

Science Notes for class 7

Class 8 Math Notes

Class 8 Chemistry Notes

Class 8 Physics Notes

Class 8 Biology Notes

Class 8 SST Notes

Class 9 Math's Notes

Class 9 Physics Notes

Class 9 Chemistry Notes

Class 9 Biology Notes

Ncert Solutions

NCERT Solutions For Class 6

NCERT Solutions For class 7

NCERT Solutions for class 8

NCERT Solutions for class 9

NCERT Solutions for class 10

NCERT Solutions for class 11

NCERT Solutions for Class 12

NCERT Solutions For Class 6

Govt Exams

Entrance Exams

Our Other Websites

Class 6th to 12th Online Courses

Class 12 Online Course

Class 11 Online Course

Class 10 Online Course

Class 9 Online Course

Class 8 Online Course

Govt Job Exams Courses

SSC Online Coaching

Bank Exam Online Coaching

TET Online Coaching

JAIIB & CAIIB Online Coaching

Bihar Exam Online Coaching

UPSC Coaching

UPSC Online Coaching

UPSC Offline & Hybrid Coaching

State PSC Online Coaching

UPPSC Online Coaching

BPSC Online Coaching

Defence Exam Coaching

NDA Online Coaching

CDS Online Coaching

AFCAT Online Coaching

Agniveer Online Coaching

Gate Exam Coaching

Civil Engineering Courses

Mechanical Engineering Courses

Formulas

Other Exams

Company Secretary Exam

Know about Physics Wallah

Physics Wallah is an Indian edtech platform that provides accessible & comprehensive learning experiences to students from Class 6th to postgraduate level. We also provide extensive NCERT solutions, sample paper, NEET, JEE Mains, BITSAT previous year papers & more such resources to students. Physics Wallah also caters to over 3.5 million registered students and over 78 lakh+ Youtube subscribers with 4.8 rating on its app.

We Stand Out because

We provide students with intensive courses with India’s qualified & experienced faculties & mentors. PW strives to make the learning experience comprehensive and accessible for students of all sections of society. We believe in empowering every single student who couldn't dream of a good career in engineering and medical field earlier.

Our Key Focus Areas

Physics Wallah's main focus is to make the learning experience as economical as possible for all students. With our affordable courses like Lakshya, Udaan and Arjuna and many others, we have been able to provide a platform for lakhs of aspirants. From providing Chemistry, Maths, Physics formula to giving e-books of eminent authors like RD Sharma, RS Aggarwal and Lakhmir Singh, PW focuses on every single student's need for preparation.

What Makes Us Different

Physics Wallah strives to develop a comprehensive pedagogical structure for students, where they get a state-of-the-art learning experience with study material and resources. Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others

These exercises are designed to strengthen understanding of AP concepts and enhance problem-solving skills through practical applications of AP formulas.

RS Aggarwal Solutions for Class 10 Maths Chapter 11 Arithmetic Progressions Exercise 11.3 Overview

RS Aggarwal Solutions for Class 10 Maths Chapter 11 Exercise 11.3 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 11 Exercise 11.3 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 11 Exercise 11.3

Here we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 11 Arithmetic Progressions Exercise 11.3 for the ease of students so that they can prepare better for their exams.

Q. The sum of the first n terms of an AP is $(3 n^{2} + 6 n) .$ Find the nth term and the 15th terms of this AP.

Solution:

Q. (i) the sum for the first n terms of an AP is $(\frac{5 n^{2}}{2} + \frac{3 n}{2})$ . Find hte nth term and the 20th term of this AP.

(ii) The sum of the first n terms of an AP is $(\frac{3 n^{2}}{2} + \frac{5 n}{2})$ . Find its nth term and the 25th term.

Solution:

(i) Sum of n terms, $S_{n} = \frac{5 n^{2}}{2} + \frac{3 n}{2}$

S_{1} = \frac{5 (1)^{2}}{2} + \frac{3 (1)}{2} = \frac{5}{2} + \frac{3}{2} = \frac{8}{2} = 4 = a

S_{1} = \frac{5 (2)^{2}}{2} + \frac{3 (2)}{2} = \frac{20}{2} + \frac{6}{2} = 10 + 3 = 13

a_{2} = S_{2} - S_{1} = 13 - 4 = 9

d = a_{2} - a_{1} = 9 - 4 = 5

a_{n} = a + (n - 1) d = 4 + (n - 1) 5 = 4 + 5 n - 5 = 5 n - 1

a_{20} = 5 n - 1 = 5 \times 20 - 1 = 100 - 1 = 99

(ii) Sum of n terms, $S_{n} = \frac{3 n^{2}}{2} + \frac{5 n}{2}$

S_{1} = \frac{3 (1)^{2}}{2} + \frac{5 (1)}{2} = \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4 = a

S_{1} = \frac{3 (2)^{2}}{2} + \frac{3 (2)}{2} = \frac{12}{2} + \frac{10}{2} = 6 + 5 = 11

a_{2} = S_{2} - S_{1} = 11 - 4 = 7

d = a_{2} - a_{1} = 7 - 4 = 3

a_{n} = a + (n - 1) d = 4 + (n - 1) 3 = 4 + 3 n - 3 = 3 n + 1

a_{25} = 3 n + 1 = 3 \times 25 + 1 = 75 + 1 = 76

Q. How many terms of the AP: 21, 18, 15, ... must be added to get the sum 0?

Solution:

Q. How many terms of the AP 9, 17, 25, ... must be taken so that their sum is 636?

Solution:

A.P = 9,17,25 a=9 d=17-9 =8 Sum of n terms,

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

636 = \frac{n}{2} [2 (9) + (n - 1) 8]

636 = \frac{n}{2} [18 + 8 n - 8]

636 = \frac{n}{2} [10 + 8 n]

1272 = 10 n + 8 n^{2}

8 n^{2} + 10 n - 1272 = 0

2 [4 n^{2} + 5 n - 636] = 0

4 n^{2} + 5 n - 636 = 0

4 n^{2} + 53 n - 48 n - 636 = 0

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

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

n - 12 = 0

n = 12

12 terms must be given to sum of 636

Q. How many terms of the AP 63, 60, 57, 54, ... must be taken so that their sum is 693. Explain the double answer.

Solution:

Given, AP 63, 60, 57 where, a = 63 and the difference (d) = 60 - 63 = -3 also given that

S_{n}

= 693 ∴ to find a, we know

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

By substituting the values of a, d and

S_{n}

we get; 693 =

\frac{n}{2}

[ 2 ×63 + (n - 1) - 3 ] 693 =

\frac{n}{2}

[ 126 - 3n + 3 ] 693 =

\frac{n}{2}

[ 129 - 3n ]

693 = \frac{129 n}{2} - \frac{3 n^{2}}{2}

693 × 2 = 129n -

3 n^{2}

1386 = 129n -

3 n^{2}

1386 - 129n +

3 n^{2}

= 0 By dividing the whole equation by 3 we get,

\frac{1386}{3} - \frac{129 n}{3} + \frac{3 n^{2}}{3} = \frac{0}{3}

462 - 43n +

n^{2}

= 0 ie;

n^{2}

- 43n + 462 = 0 using factorisation method :-- sum = - 43 and product = 462 ∴ the numbers are -21 and -22 So by splitting the middle term we get; (

n^{2}

S_{21} = \frac{21}{2} (a + a_{21})

a_{21}

= a + ( 21 - 1 ) d = 63 + [ 20 × ( - 3 ) ] = 63 - 60

a_{21} = 3

∴

S_{21} = \frac{21}{2}

[ 63 + 3 ] =

\frac{21}{2}

× 66

S_{21}

= 693 So the condition is satisfied for when n = 21 Now check for when n = 22

S_{22}

\frac{22}{2}

( a +

a_{22}

)

a_{22}

= a + ( 22 - 1 ) d = 63 + [ 21 × ( -3 ) ] = 63 - 63

a_{22}

= 0 We know

S_{22}

S_{21}

a_{22}

= 693 + 0 = 693 ∴ the condition is satisfied in both the cases so n = 21 or n = 22 Q. Find the sum of all odd numbers between 0 and 50.

Solution:

1st odd no.=1 last odd no.=49 a=1, an or l=49, common diff. d= 2,

a_{n} = a + (n - 1) d

=1+(n-1)2=49 =1+2n-2=49 =2n-1=49 =2n=49=1 =2n=50 n=25

S_{n} = \frac{n}{2} [a + L]

= \frac{25}{2} (1 + 49)

= \frac{25}{2} \times 50

= 25 \times 25 = 625

therefore sum of all odd integers between 0 to 50 is 625

Q. Find the sum of all natural numbers between 200 and 400 which are divisible by 7.

Solution:

a_{n} = a + (n - 1) d

= 399 ⇒ 399 = 203 + (n –1) 7 ⇒ 7 (n –1) = 196 ⇒ n –1 = 28 ⇒ n = 29

S_{n} = \frac{n}{2} [a + L]

S_{29} = \frac{29}{2} [203 + 399]

= \frac{29}{2} \times 602 = 8729

Thus, the required sum is 8729.

Q. Find the sum first forty positive integers divisible by 6.

Solution:

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

= \frac{40}{2} \times 2 (6) + (39) 6

= 20 12 + 234 = 20 \times 246 = 4920

Q. Find the sum of the first 15 multiples of 8.

Solution:

The first 8 multiples of 8 are 8, 16, 24, 32, 40, 48, 56,64.......120 These are in an A.P., having first term as 8 and common difference as 8. Therefore, a = 8 d = 8

S_{15}

= ?

S_{n}

\frac{n}{2}

[2a + (n - 1)d]

\Rightarrow

S_{15}

\frac{15}{2}

[2(8) + (15 - 1)8]

\Rightarrow

S_{15}

= 7.5[6 + (14) (8)]

\Rightarrow

S_{15}

= 7.5[16 + 112]

\Rightarrow

S_{15}

= 7.5(128)

\Rightarrow

S_{15}

= 960

Q. Find the sum of all three - digit natural numbers which are divisible by 13.

Solution:

As per question; AP: 104 , 117 , 130 ........... 988 a = 104 ,

a_{n}

= 988

\Rightarrow

988 = 104 + (n-1) 13

\Rightarrow

988 - 104 = (n-1)13

\Rightarrow

\frac{884}{13}

= n-1

\Rightarrow

n= 69

S_{n}

\frac{n}{2}

(

a + a_{n}

)

\Rightarrow

S_{n}

\frac{n}{2}

( 104 + 988)

\Rightarrow

S_{n}

\frac{69}{2}

( 1092 )

\Rightarrow

S_{n}

= 37674

Q. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Solution:

Let a and d are the first term and the common difference for an AP respectively. Number of terms of AP = n Last term = nth term = l =

a_{n}

a = 17 , d = 9 , l = 350 a + ( n - 1 ) d = 350

\Rightarrow

17 + ( n - 1 ) 9 = 350

\Rightarrow

( n - 1 ) 9 = 350 - 17

\Rightarrow

( n - 1 ) 9 = 333

\Rightarrow

n - 1 =

\frac{333}{9}

\Rightarrow

n - 1 = 37

\Rightarrow

n = 37 + 1

\Rightarrow

n = 38 Therefore, number of terms in given AP = n = 38 Sum of n terms of AP =

S_{n}

S_{n}

( a + l ) Here, n= 38

\Rightarrow

S_{38}

= 19[ 17 + 350 ]

\Rightarrow

S_{38}

= 19 × 367

\Rightarrow

S_{38}

= 6973

Q. In an AP, the first term is 22, nth term is -11 and sum of first n terms is 66. Find n and hence find the common difference.

Solution:

Given: The number of trees that each section of each class will plant will be double of the class in which they are studying. Number of classes are

12

(

1

12

), and each class has two sections.

\Rightarrow

Number of trees planted by the students of each section of class

1 = 1 \times 2 = 2

There are two sections of class

1

. Therefore, Number of trees planted by the students of class

1 = 2 \times 2 = 4

Number of trees planted by the students of each section of class

2 = 2 \times 2 = 4

.---(1) There are two sections of class

2

. Therefore, Number of trees planted by the students of class

2 = 2 \times 4 = 8

---(2) Similarly, Number of trees planted by the students of class

3 = 2 \times 6 = 12

---(3) So, the number of trees planted by the students of different classes are

4, 8, 12, . . .

Therefore, Total number of trees planted by the students

= 4 + 8 + 12 + . . . .

up to

12

terms Lets evaluate,

4 + 8 + 12 + . . . .

up to

12

terms. Here,

8 - 4 = 12 - 8 = . . . . = 4

Since, the difference of every consecutive terms is constant. So, this series is an arithmetic series. Here,

a = 4, d = 8 - 4 = 4

and

n = 12

Since, the sum of first

n

- terms of an AP is,,

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

\Rightarrow S_{12} = \frac{12}{2} [2 \times 4 + (12 - 1) \times 4]

\Rightarrow S_{12} = 6 [8 + (11 \times 4)]

\Rightarrow S_{12} = 6 \times [8 + 44]

\Rightarrow S_{12} = 6 \times 52

\Rightarrow S_{12} = 312

Therefore, the total number of trees planted by the students are

312

Solution:

S_{10} = ? S_{10} = \frac{12}{2} [2 (10) + (n - 1) (6)] = \frac{12}{2} [2 (10) + (10 - 1) (6)] = 5 [20 + 54] = 5 (74) = 370

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

Solution:

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

we get

S_{25} = \frac{25}{2} [2 \times 20 + (25 - 1) \times 10]

= \frac{25}{2} \times (40 + 240)

= \frac{25}{2} \times 280 = 3500

Hence, the total distance covered by the gardener to water all the trees is 3500 m.

Q. The 24th term of an AP is twice is 10th term. Show that its 72nd term is 4 times its 15 the term.

Solution:

Let first term = a and common difference = d we know that

a_{n} = a + (n - 1) d

Given that,

a_{24} = 2 \times a_{10}

\Rightarrow a + (24 - 1) d = 2 \times (a + (10 - 1) d)

\Rightarrow a + 23 d = 2 \times (a + 9 d)

\Rightarrow a + 23 d = 2 a + 18 d)

\Rightarrow a = 5 d . . . . (i)

Now 15th term,

a_{15} = a + (15 - 1) d = a + 14 d [From (i)] = 5 d + 14 d = 19 d

Similarly 72nd term,

a_{72} = a + (72 - 1) d = a + 71 d = 5 d + 71 d = 76 d

Now,

\frac{a_{72}}{a_{15}} = \frac{76 d}{19 d} = 4

∴ a_{72} = 4 \times a_{15}

Hence,

72^{t h}

term is

4

times to its

15^{t h}

term.

Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 11 Exercise 11.3

Clear Understanding: These solutions provide step-by-step explanations that clarify how to solve problems involving AP, helping students grasp concepts effectively.

Practice: By solving problems from Exercise 11.3 students can practice applying AP formulas and strengthen their problem-solving skills.

Comprehensive Coverage: The solutions cover various types of problems ensuring students are well-prepared for exams and assessments.

Self-assessment: Students can use these solutions to self-assess their progress and identify areas where they may need further practice or clarification.