Open App

Class 10 Exam > Class 10 Notes > Mathematics (Maths) Class 10 > RD Sharma Solutions: Arithmetic Progressions (Exercise 5D)

Arithmetic Progressions (Exercise 5D) RD Sharma Solutions | Mathematics (Maths) Class 10 PDF Download

Download, print and study this document offline

Please wait while the PDF view is loading

 Page 1


 
1. Find the sum of each of the following Aps:
(iii) -37, -33, -
(iv) 
1 1 1
, , ,.....
15 12 10
to 11 terms. 
Sol:
(i)
Here, 2 a and 7 2 5 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
19
19
2 2 19 1 5
2
19
4 90
2
19
94
2
893
S
(ii)
Here, 9 a and 7 9 2 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
14
14
2 9 14 1 2
2
7 18 26
7 8
56
S
(iii) The given AP is -37, -33, -
Here, 37 a and 33 37 33 37 4 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
12
12
2 37 12 1 4
2
S
6 74 44
6 30
Page 2


 
1. Find the sum of each of the following Aps:
(iii) -37, -33, -
(iv) 
1 1 1
, , ,.....
15 12 10
to 11 terms. 
Sol:
(i)
Here, 2 a and 7 2 5 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
19
19
2 2 19 1 5
2
19
4 90
2
19
94
2
893
S
(ii)
Here, 9 a and 7 9 2 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
14
14
2 9 14 1 2
2
7 18 26
7 8
56
S
(iii) The given AP is -37, -33, -
Here, 37 a and 33 37 33 37 4 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
12
12
2 37 12 1 4
2
S
6 74 44
6 30
 
180
(iv) The given AP is
1 1 1
, , ,......
15 12 10
Here, 
1
15
a and 
1 1 5 4 1
12 15 60 60
d
Using the formula, 2 1 ,
2
n
n
S a n d we have
11
11 1 1
2 11 1
2 15 60
11 2 10
2 15 60
S
11 18
2 60
33
20
(v) The given AP is 0.6,1.7,2.8,..........
Here, 0.6 a and 1.7 0.6 1.1 d
Using formula, 2 1 ,
2
n
n
S a n d we have
100
100
2 0.6 100 1 1.1
2
S
50 1.2 108.9
50 110.1
5505
2. Find the sum of each of the following arithmetic series:
(i) 
1
7 10 14 ... 84
2
(ii) 34 32 30 ... 10
(iii) ( 5) ( 8) ( 11) ... ( 230)
Sol:
(i) The given arithmetic series is 
1
7 10 14 ..... 84.
2
Here, 
1 21 21 4 7
7, 10 7 7
2 2 2 2
a d and 84. l
Let the given series contains n terms. Then,
Page 3


 
1. Find the sum of each of the following Aps:
(iii) -37, -33, -
(iv) 
1 1 1
, , ,.....
15 12 10
to 11 terms. 
Sol:
(i)
Here, 2 a and 7 2 5 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
19
19
2 2 19 1 5
2
19
4 90
2
19
94
2
893
S
(ii)
Here, 9 a and 7 9 2 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
14
14
2 9 14 1 2
2
7 18 26
7 8
56
S
(iii) The given AP is -37, -33, -
Here, 37 a and 33 37 33 37 4 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
12
12
2 37 12 1 4
2
S
6 74 44
6 30
 
180
(iv) The given AP is
1 1 1
, , ,......
15 12 10
Here, 
1
15
a and 
1 1 5 4 1
12 15 60 60
d
Using the formula, 2 1 ,
2
n
n
S a n d we have
11
11 1 1
2 11 1
2 15 60
11 2 10
2 15 60
S
11 18
2 60
33
20
(v) The given AP is 0.6,1.7,2.8,..........
Here, 0.6 a and 1.7 0.6 1.1 d
Using formula, 2 1 ,
2
n
n
S a n d we have
100
100
2 0.6 100 1 1.1
2
S
50 1.2 108.9
50 110.1
5505
2. Find the sum of each of the following arithmetic series:
(i) 
1
7 10 14 ... 84
2
(ii) 34 32 30 ... 10
(iii) ( 5) ( 8) ( 11) ... ( 230)
Sol:
(i) The given arithmetic series is 
1
7 10 14 ..... 84.
2
Here, 
1 21 21 4 7
7, 10 7 7
2 2 2 2
a d and 84. l
Let the given series contains n terms. Then,
 
84
7
7 1 84 1
2
7 7
84
2 2
7 7 161
84
2 2 2
161
23
7
n
n
a
n a a n d
n
n
n
Required sum 
23
7 84
2 2
n
n
S a l
23
91
2
2030
2
1
1046
2
(ii) The given arithmetic series is 34 32 30 ..... 10.
Here, 34, 32 34 2 a d and 10. l
Let the given series contain n terms. Then,
10
34 1 2 10 1
2 36 10
2 10 36 26
13
n
n
a
n a a n d
n
n
n
Required sum 
13
34 10
2 2
n
n
S a l
13
44
2
286
(iii) The given arithmetic series is 5 8 11 ....... 230 .
Here, 5, 8 5 8 5 3 a d and 230. l
Let the given series contain n terms. Then,
Page 4


 
1. Find the sum of each of the following Aps:
(iii) -37, -33, -
(iv) 
1 1 1
, , ,.....
15 12 10
to 11 terms. 
Sol:
(i)
Here, 2 a and 7 2 5 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
19
19
2 2 19 1 5
2
19
4 90
2
19
94
2
893
S
(ii)
Here, 9 a and 7 9 2 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
14
14
2 9 14 1 2
2
7 18 26
7 8
56
S
(iii) The given AP is -37, -33, -
Here, 37 a and 33 37 33 37 4 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
12
12
2 37 12 1 4
2
S
6 74 44
6 30
 
180
(iv) The given AP is
1 1 1
, , ,......
15 12 10
Here, 
1
15
a and 
1 1 5 4 1
12 15 60 60
d
Using the formula, 2 1 ,
2
n
n
S a n d we have
11
11 1 1
2 11 1
2 15 60
11 2 10
2 15 60
S
11 18
2 60
33
20
(v) The given AP is 0.6,1.7,2.8,..........
Here, 0.6 a and 1.7 0.6 1.1 d
Using formula, 2 1 ,
2
n
n
S a n d we have
100
100
2 0.6 100 1 1.1
2
S
50 1.2 108.9
50 110.1
5505
2. Find the sum of each of the following arithmetic series:
(i) 
1
7 10 14 ... 84
2
(ii) 34 32 30 ... 10
(iii) ( 5) ( 8) ( 11) ... ( 230)
Sol:
(i) The given arithmetic series is 
1
7 10 14 ..... 84.
2
Here, 
1 21 21 4 7
7, 10 7 7
2 2 2 2
a d and 84. l
Let the given series contains n terms. Then,
 
84
7
7 1 84 1
2
7 7
84
2 2
7 7 161
84
2 2 2
161
23
7
n
n
a
n a a n d
n
n
n
Required sum 
23
7 84
2 2
n
n
S a l
23
91
2
2030
2
1
1046
2
(ii) The given arithmetic series is 34 32 30 ..... 10.
Here, 34, 32 34 2 a d and 10. l
Let the given series contain n terms. Then,
10
34 1 2 10 1
2 36 10
2 10 36 26
13
n
n
a
n a a n d
n
n
n
Required sum 
13
34 10
2 2
n
n
S a l
13
44
2
286
(iii) The given arithmetic series is 5 8 11 ....... 230 .
Here, 5, 8 5 8 5 3 a d and 230. l
Let the given series contain n terms. Then,
 
230
5 1 3 230 1
3 2 230
3 230 2 228
76
n
n
a
n a a n d
n
n
n
Required sum
76
5 230
2 2
n
n
S a l
76
235
2
8930
3. Find the sum of first n terms of an AP whose nth term is (5 - 6n). Hence, find the sum of its 
first 20 terms.
Sol:
Let 
n
a be the nth term of the AP.
5 6
n
a n
Putting 1, n we get
First term,
1
5 6 1 1 a a
Putting 2, n we get
2
5 6 2 7 a
Let d be the common difference of the AP.
2 1
7 ( 1) 7 1 6 d a a
Sum of first n tern of the AP, 
n
S
2
2 1 1 6 2 1
2 2
2 6 6
2
2 3
2 3
n
n n
n S a n d
n
n
n n
n n
Putting 20, n we get
2
20
2 20 3 20 40 1200 1160 S
4. The sum of the first n terms of an AP is 
2
3 6 . n n Find the nth term and the 15
th
 term of this 
AP.
Sol:
Page 5


 
1. Find the sum of each of the following Aps:
(iii) -37, -33, -
(iv) 
1 1 1
, , ,.....
15 12 10
to 11 terms. 
Sol:
(i)
Here, 2 a and 7 2 5 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
19
19
2 2 19 1 5
2
19
4 90
2
19
94
2
893
S
(ii)
Here, 9 a and 7 9 2 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
14
14
2 9 14 1 2
2
7 18 26
7 8
56
S
(iii) The given AP is -37, -33, -
Here, 37 a and 33 37 33 37 4 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
12
12
2 37 12 1 4
2
S
6 74 44
6 30
 
180
(iv) The given AP is
1 1 1
, , ,......
15 12 10
Here, 
1
15
a and 
1 1 5 4 1
12 15 60 60
d
Using the formula, 2 1 ,
2
n
n
S a n d we have
11
11 1 1
2 11 1
2 15 60
11 2 10
2 15 60
S
11 18
2 60
33
20
(v) The given AP is 0.6,1.7,2.8,..........
Here, 0.6 a and 1.7 0.6 1.1 d
Using formula, 2 1 ,
2
n
n
S a n d we have
100
100
2 0.6 100 1 1.1
2
S
50 1.2 108.9
50 110.1
5505
2. Find the sum of each of the following arithmetic series:
(i) 
1
7 10 14 ... 84
2
(ii) 34 32 30 ... 10
(iii) ( 5) ( 8) ( 11) ... ( 230)
Sol:
(i) The given arithmetic series is 
1
7 10 14 ..... 84.
2
Here, 
1 21 21 4 7
7, 10 7 7
2 2 2 2
a d and 84. l
Let the given series contains n terms. Then,
 
84
7
7 1 84 1
2
7 7
84
2 2
7 7 161
84
2 2 2
161
23
7
n
n
a
n a a n d
n
n
n
Required sum 
23
7 84
2 2
n
n
S a l
23
91
2
2030
2
1
1046
2
(ii) The given arithmetic series is 34 32 30 ..... 10.
Here, 34, 32 34 2 a d and 10. l
Let the given series contain n terms. Then,
10
34 1 2 10 1
2 36 10
2 10 36 26
13
n
n
a
n a a n d
n
n
n
Required sum 
13
34 10
2 2
n
n
S a l
13
44
2
286
(iii) The given arithmetic series is 5 8 11 ....... 230 .
Here, 5, 8 5 8 5 3 a d and 230. l
Let the given series contain n terms. Then,
 
230
5 1 3 230 1
3 2 230
3 230 2 228
76
n
n
a
n a a n d
n
n
n
Required sum
76
5 230
2 2
n
n
S a l
76
235
2
8930
3. Find the sum of first n terms of an AP whose nth term is (5 - 6n). Hence, find the sum of its 
first 20 terms.
Sol:
Let 
n
a be the nth term of the AP.
5 6
n
a n
Putting 1, n we get
First term,
1
5 6 1 1 a a
Putting 2, n we get
2
5 6 2 7 a
Let d be the common difference of the AP.
2 1
7 ( 1) 7 1 6 d a a
Sum of first n tern of the AP, 
n
S
2
2 1 1 6 2 1
2 2
2 6 6
2
2 3
2 3
n
n n
n S a n d
n
n
n n
n n
Putting 20, n we get
2
20
2 20 3 20 40 1200 1160 S
4. The sum of the first n terms of an AP is 
2
3 6 . n n Find the nth term and the 15
th
 term of this 
AP.
Sol:
 
Let 
n
S denotes the sum of first n terms of the AP.
2
2
1
2
2
3 6
3 1 6 1
3 2 1 6 1
3 3
n
n
S n n
S n n
n n n
n
t h
n term of the AP, 
n
a
1
2 2
3 6 3 3
6 3
n n
S S
n n n
n
Putting 15, n we get
15
6 15 3 90 3 93 a
Hence, the 
t h
n term is 6 3 n and 15
th
term is 93.
5. The sum of the  first n terms of an AP is given by 
2
3
n
S n n . Find its 
(i) nth term, 
(ii) first term and  
(iii) common difference.
Sol:
Given: 
2
3 ......
n
S n n i
Replacing n by 1 n in (i), we get:
2
1
2
2
3 1 1
3 2 1 1
3 7 4
n
S n n
n n n
n n
(i) Now, 
1 n n n
T S S
2 2
3 3 7 4 6 4 n n n n n
t h
n term, 6 4 .......
n
T n i i
(ii) Putting 1 n in (ii), we get:
1
6 1 4 2 T
(iii) Putting 2 n in (ii), we get:
2
6 2 4 8 T
Common difference, 
2 1
8 2 6 d T T

	Mathematics (Maths) Class 10 126 videos\|457 docs\|75 tests

Mathematics (Maths) Class 10

126 videos|457 docs|75 tests

Join Course for Free

Top Courses for Class 10

View all

Related Exams

Class 10

About this Document

	4.85/5 Rating
	Dec 23, 2024 Last updated

Document Description: RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The notes and questions for RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) have been prepared according to the Class 10 exam syllabus. Information about RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) covers topics like and RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) Example, for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Arithmetic Progressions (Exercise 5D).

Introduction of RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) in English is available as part of our Mathematics (Maths) Class 10 for Class 10 & RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) in Hindi for Mathematics (Maths) Class 10 course. Download more important topics related with notes, lectures and mock test series for Class 10 Exam by signing up for free. Class 10: Arithmetic Progressions (Exercise 5D) RD Sharma Solutions | Mathematics (Maths) Class 10

Description

Full syllabus notes, lecture & questions for Arithmetic Progressions (Exercise 5D) RD Sharma Solutions | Mathematics (Maths) Class 10 - Class 10 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 10 | Best notes, free PDF download

Information about RD Sharma Solutions: Arithmetic Progressions (Exercise 5D)

In this doc you can find the meaning of RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) defined & explained in the simplest way possible. Besides explaining types of RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) tests, examples and also practice Class 10 tests

	Mathematics (Maths) Class 10 126 videos\|457 docs\|75 tests

Mathematics (Maths) Class 10

126 videos|457 docs|75 tests

Join Course for Free

Download as PDF

Explore Courses for Class 10 exam

Top Courses for Class 10

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

Viva Questions

practice quizzes

Previous Year Questions with Solutions

Summary

study material

Important questions

Free

Arithmetic Progressions (Exercise 5D) RD Sharma Solutions | Mathematics (Maths) Class 10

shortcuts and tricks

Objective type Questions

ppt

mock tests for examination

Arithmetic Progressions (Exercise 5D) RD Sharma Solutions | Mathematics (Maths) Class 10

Extra Questions

Semester Notes

Sample Paper

past year papers

pdf

MCQs

video lectures

Exam

Arithmetic Progressions (Exercise 5D) RD Sharma Solutions | Mathematics (Maths) Class 10

;

Additional Information about RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) for Class 10 Preparation

RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) Free PDF Download

The RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) is an invaluable resource that delves deep into the core of the Class 10 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) now and kickstart your journey towards success in the Class 10 exam.

Importance of RD Sharma Solutions: Arithmetic Progressions (Exercise 5D)

The importance of RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) cannot be overstated, especially for Class 10 aspirants. This document holds the key to success in the Class 10 exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) Notes

RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to RD Sharma Solutions: Arithmetic Progressions (Exercise 5D). It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) Notes on EduRev are your ultimate resource for success.

RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) Class 10 Questions

The "RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) Class 10 Questions" guide is a valuable resource for all aspiring students preparing for the Class 10 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) on the App

Students of Class 10 can study RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Arithmetic Progressions (Exercise 5D), students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) is prepared as per the latest Class 10 syllabus.

Education Revolution