Class 10 Exam > Class 10 Notes > Mathematics (Maths) Class 10 > RD Sharma Solutions: Arithmetic Progressions (Exercise 5A)
Arithmetic Progressions (Exercise 5A) 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. Show that each of the progressions given below is an AP. Find the first term, common difference and next term of each. (i) (ii) 11, 6, 1, (iii) 5 2 1 1, , , ,.......... 6 3 2 (iv) 2, 8, 18, 32,.......... (v) 20, 45, 80, 125,......... Sol: (i) The given progression 9 Clearly, 15 9 = 21 15 = 27 21 = 6 (Constant) Thus, each term differs from its preceding term by 6. So, the given progression is an AP. First term = 9 Common difference =6 Next term of the AP = 27 + 6 = 33 (ii) The given progression 11, 6, 1, Clearly,6 11 = 1 6 = 4 1 = 5 (Constant) Thus, each term differs from its preceding term by 6. So, the given progression is an AP. First term = 11 Common difference = 5 Next term of the 4 5 9 A P (iii) The given progression 5 2 1 1, , , ,.......... 6 3 2 Clearly, 5 2 5 1 2 1 1 6 3 6 2 3 6 (Constant) Thus, each term differs from its preceding term by 1 . 6 So, the given progression is an AP. First term = 1 Common difference = 1 6 Next tern of the 1 1 2 1 2 6 6 3 A P Page 2 1. Show that each of the progressions given below is an AP. Find the first term, common difference and next term of each. (i) (ii) 11, 6, 1, (iii) 5 2 1 1, , , ,.......... 6 3 2 (iv) 2, 8, 18, 32,.......... (v) 20, 45, 80, 125,......... Sol: (i) The given progression 9 Clearly, 15 9 = 21 15 = 27 21 = 6 (Constant) Thus, each term differs from its preceding term by 6. So, the given progression is an AP. First term = 9 Common difference =6 Next term of the AP = 27 + 6 = 33 (ii) The given progression 11, 6, 1, Clearly,6 11 = 1 6 = 4 1 = 5 (Constant) Thus, each term differs from its preceding term by 6. So, the given progression is an AP. First term = 11 Common difference = 5 Next term of the 4 5 9 A P (iii) The given progression 5 2 1 1, , , ,.......... 6 3 2 Clearly, 5 2 5 1 2 1 1 6 3 6 2 3 6 (Constant) Thus, each term differs from its preceding term by 1 . 6 So, the given progression is an AP. First term = 1 Common difference = 1 6 Next tern of the 1 1 2 1 2 6 6 3 A P (iv) The given progression 2, 8, 18, 32,.......... This sequence can be written as 2, 2 2,3 2,4 2,........ Clearly, 2 2 2 3 2 2 2 4 2 3 2 2 (Constant) Thus, each term differs from its preceding term by 2, So, the given progression is an AP. First term 2 Common difference 2 Next tern of the 4 2 2 5 2 50 A P (v) This given progression 20, 45, 80, 125,......... This sequence can be re-written as 2 5,3 5, 4 5,5 5,........... Clearly, 3 5 2 5 4 5 3 5 5 5 4 5 5 (Constant) Thus, each term differs from its preceding term by 5.So, the given progression is an AP. First term 2 5 20 Common difference 5 Next term of the 5 5 5 6 5 180 A P 2. Find: (i) the 20 th term of the AP 9,13,17,21,.......... (ii) the 35 th term of AP 20,17,14,11,.......... (iii) the 18 th term of the AP 2, 18, 50, 98,........... (iv) the 9 th term of the AP 3 5 7 9 , , , ,......... 4 4 4 4 (v) the 15the term of the AP 40, 15,10,35,......... Sol: (i) The given A P is 9,13,17,21,.......... First term, 9 a Common difference, 13 9 4 d t h n term of the AP, 1 9 1 4 n a a n d n 20 t h term of the AP, 20 9 20 1 4 9 76 85 a (ii) The given AP is 20,17,14,11,.......... First term, 20 a Common difference, 17 20 3 d t h n term of the AP, 1 20 1 3 n a a n d n Page 3 1. Show that each of the progressions given below is an AP. Find the first term, common difference and next term of each. (i) (ii) 11, 6, 1, (iii) 5 2 1 1, , , ,.......... 6 3 2 (iv) 2, 8, 18, 32,.......... (v) 20, 45, 80, 125,......... Sol: (i) The given progression 9 Clearly, 15 9 = 21 15 = 27 21 = 6 (Constant) Thus, each term differs from its preceding term by 6. So, the given progression is an AP. First term = 9 Common difference =6 Next term of the AP = 27 + 6 = 33 (ii) The given progression 11, 6, 1, Clearly,6 11 = 1 6 = 4 1 = 5 (Constant) Thus, each term differs from its preceding term by 6. So, the given progression is an AP. First term = 11 Common difference = 5 Next term of the 4 5 9 A P (iii) The given progression 5 2 1 1, , , ,.......... 6 3 2 Clearly, 5 2 5 1 2 1 1 6 3 6 2 3 6 (Constant) Thus, each term differs from its preceding term by 1 . 6 So, the given progression is an AP. First term = 1 Common difference = 1 6 Next tern of the 1 1 2 1 2 6 6 3 A P (iv) The given progression 2, 8, 18, 32,.......... This sequence can be written as 2, 2 2,3 2,4 2,........ Clearly, 2 2 2 3 2 2 2 4 2 3 2 2 (Constant) Thus, each term differs from its preceding term by 2, So, the given progression is an AP. First term 2 Common difference 2 Next tern of the 4 2 2 5 2 50 A P (v) This given progression 20, 45, 80, 125,......... This sequence can be re-written as 2 5,3 5, 4 5,5 5,........... Clearly, 3 5 2 5 4 5 3 5 5 5 4 5 5 (Constant) Thus, each term differs from its preceding term by 5.So, the given progression is an AP. First term 2 5 20 Common difference 5 Next term of the 5 5 5 6 5 180 A P 2. Find: (i) the 20 th term of the AP 9,13,17,21,.......... (ii) the 35 th term of AP 20,17,14,11,.......... (iii) the 18 th term of the AP 2, 18, 50, 98,........... (iv) the 9 th term of the AP 3 5 7 9 , , , ,......... 4 4 4 4 (v) the 15the term of the AP 40, 15,10,35,......... Sol: (i) The given A P is 9,13,17,21,.......... First term, 9 a Common difference, 13 9 4 d t h n term of the AP, 1 9 1 4 n a a n d n 20 t h term of the AP, 20 9 20 1 4 9 76 85 a (ii) The given AP is 20,17,14,11,.......... First term, 20 a Common difference, 17 20 3 d t h n term of the AP, 1 20 1 3 n a a n d n 35 t h term of the AP, 35 20 35 1 3 20 102 82 a (iii) The given AP is 2, 18, 50, 98,........... This can be re-written as 2,3 2,5 2,7 2,......... First term, 2 a Common difference, 3 2 2 2 2 d t h n term of the AP, 1 2 1 2 2 n a a n d n 18 t h term of the AP, 18 2 18 1 2 2 2 34 2 35 2 2450 a (iv) The given AP is 3 5 7 9 , , , ,......... 4 4 4 4 First term, 3 4 a Common difference, 5 3 2 1 4 4 4 2 d t h n term of the AP, 3 1 1 1 4 2 n a a n d n 9 t h t e r m of the AP, 9 3 1 3 19 9 1 4 4 2 4 4 a (v) The given AP is 40, 15,10,35,......... First term, 40 a Common difference, 15 40 25 d t h n term of the AP, 1 40 1 25 n a a n d n 15 t h term of the AP, 15 40 15 1 25 40 350 310 a 3. Find the 37 th term of the AP 3 1 1 6,7 ,9 ,11 ,......... 4 2 4 Sol: The given AP is 3 1 1 6,7 ,9 ,11 ,......... 4 2 4 First term, 6 a and common difference, 3 31 31 24 7 7 6 6 4 4 4 4 d Now, 37 37 1 36 T a d a d 7 6 36 6 63 69 4 37 t h term = 69 Page 4 1. Show that each of the progressions given below is an AP. Find the first term, common difference and next term of each. (i) (ii) 11, 6, 1, (iii) 5 2 1 1, , , ,.......... 6 3 2 (iv) 2, 8, 18, 32,.......... (v) 20, 45, 80, 125,......... Sol: (i) The given progression 9 Clearly, 15 9 = 21 15 = 27 21 = 6 (Constant) Thus, each term differs from its preceding term by 6. So, the given progression is an AP. First term = 9 Common difference =6 Next term of the AP = 27 + 6 = 33 (ii) The given progression 11, 6, 1, Clearly,6 11 = 1 6 = 4 1 = 5 (Constant) Thus, each term differs from its preceding term by 6. So, the given progression is an AP. First term = 11 Common difference = 5 Next term of the 4 5 9 A P (iii) The given progression 5 2 1 1, , , ,.......... 6 3 2 Clearly, 5 2 5 1 2 1 1 6 3 6 2 3 6 (Constant) Thus, each term differs from its preceding term by 1 . 6 So, the given progression is an AP. First term = 1 Common difference = 1 6 Next tern of the 1 1 2 1 2 6 6 3 A P (iv) The given progression 2, 8, 18, 32,.......... This sequence can be written as 2, 2 2,3 2,4 2,........ Clearly, 2 2 2 3 2 2 2 4 2 3 2 2 (Constant) Thus, each term differs from its preceding term by 2, So, the given progression is an AP. First term 2 Common difference 2 Next tern of the 4 2 2 5 2 50 A P (v) This given progression 20, 45, 80, 125,......... This sequence can be re-written as 2 5,3 5, 4 5,5 5,........... Clearly, 3 5 2 5 4 5 3 5 5 5 4 5 5 (Constant) Thus, each term differs from its preceding term by 5.So, the given progression is an AP. First term 2 5 20 Common difference 5 Next term of the 5 5 5 6 5 180 A P 2. Find: (i) the 20 th term of the AP 9,13,17,21,.......... (ii) the 35 th term of AP 20,17,14,11,.......... (iii) the 18 th term of the AP 2, 18, 50, 98,........... (iv) the 9 th term of the AP 3 5 7 9 , , , ,......... 4 4 4 4 (v) the 15the term of the AP 40, 15,10,35,......... Sol: (i) The given A P is 9,13,17,21,.......... First term, 9 a Common difference, 13 9 4 d t h n term of the AP, 1 9 1 4 n a a n d n 20 t h term of the AP, 20 9 20 1 4 9 76 85 a (ii) The given AP is 20,17,14,11,.......... First term, 20 a Common difference, 17 20 3 d t h n term of the AP, 1 20 1 3 n a a n d n 35 t h term of the AP, 35 20 35 1 3 20 102 82 a (iii) The given AP is 2, 18, 50, 98,........... This can be re-written as 2,3 2,5 2,7 2,......... First term, 2 a Common difference, 3 2 2 2 2 d t h n term of the AP, 1 2 1 2 2 n a a n d n 18 t h term of the AP, 18 2 18 1 2 2 2 34 2 35 2 2450 a (iv) The given AP is 3 5 7 9 , , , ,......... 4 4 4 4 First term, 3 4 a Common difference, 5 3 2 1 4 4 4 2 d t h n term of the AP, 3 1 1 1 4 2 n a a n d n 9 t h t e r m of the AP, 9 3 1 3 19 9 1 4 4 2 4 4 a (v) The given AP is 40, 15,10,35,......... First term, 40 a Common difference, 15 40 25 d t h n term of the AP, 1 40 1 25 n a a n d n 15 t h term of the AP, 15 40 15 1 25 40 350 310 a 3. Find the 37 th term of the AP 3 1 1 6,7 ,9 ,11 ,......... 4 2 4 Sol: The given AP is 3 1 1 6,7 ,9 ,11 ,......... 4 2 4 First term, 6 a and common difference, 3 31 31 24 7 7 6 6 4 4 4 4 d Now, 37 37 1 36 T a d a d 7 6 36 6 63 69 4 37 t h term = 69 4. Find the 25 th term of the AP 1 1 5,4 , 4,3 ,3,........ 2 2 Sol: The given AP is 1 1 5,4 , 4,3 ,3,........ 2 2 First term = 5 Common difference 1 9 9 10 1 4 5 5 2 2 2 2 1 5 2 a a n d d Now, 25 25 1 24 T a d a d 1 5 24 5 12 7 2 25 t h term 7 5. Find the nth term of each of the following Aps: (ii) 16, 9, 2, - Sol: (i) (6n 1) (ii) (23 7n) 6. If the nth term of a progression is (4n 10) show that it is an AP. Find its (i) first term, (ii) common difference (iii) 16 the term. Sol: 4 10 n n T [Given] 1 4 1 10 6 T 2 3 4 4 2 10 2 4 3 10 2 4 4 10 6 T T T Clearly, 2 6 2 2 6 2 4 (Constant) So, the terms 6, 2,2,6,...... forms an AP. Thus we have (i) First term 6 (ii) Common difference = 4 (iii) 16 1 15 6 15 4 54 T a n d a d Page 5 1. Show that each of the progressions given below is an AP. Find the first term, common difference and next term of each. (i) (ii) 11, 6, 1, (iii) 5 2 1 1, , , ,.......... 6 3 2 (iv) 2, 8, 18, 32,.......... (v) 20, 45, 80, 125,......... Sol: (i) The given progression 9 Clearly, 15 9 = 21 15 = 27 21 = 6 (Constant) Thus, each term differs from its preceding term by 6. So, the given progression is an AP. First term = 9 Common difference =6 Next term of the AP = 27 + 6 = 33 (ii) The given progression 11, 6, 1, Clearly,6 11 = 1 6 = 4 1 = 5 (Constant) Thus, each term differs from its preceding term by 6. So, the given progression is an AP. First term = 11 Common difference = 5 Next term of the 4 5 9 A P (iii) The given progression 5 2 1 1, , , ,.......... 6 3 2 Clearly, 5 2 5 1 2 1 1 6 3 6 2 3 6 (Constant) Thus, each term differs from its preceding term by 1 . 6 So, the given progression is an AP. First term = 1 Common difference = 1 6 Next tern of the 1 1 2 1 2 6 6 3 A P (iv) The given progression 2, 8, 18, 32,.......... This sequence can be written as 2, 2 2,3 2,4 2,........ Clearly, 2 2 2 3 2 2 2 4 2 3 2 2 (Constant) Thus, each term differs from its preceding term by 2, So, the given progression is an AP. First term 2 Common difference 2 Next tern of the 4 2 2 5 2 50 A P (v) This given progression 20, 45, 80, 125,......... This sequence can be re-written as 2 5,3 5, 4 5,5 5,........... Clearly, 3 5 2 5 4 5 3 5 5 5 4 5 5 (Constant) Thus, each term differs from its preceding term by 5.So, the given progression is an AP. First term 2 5 20 Common difference 5 Next term of the 5 5 5 6 5 180 A P 2. Find: (i) the 20 th term of the AP 9,13,17,21,.......... (ii) the 35 th term of AP 20,17,14,11,.......... (iii) the 18 th term of the AP 2, 18, 50, 98,........... (iv) the 9 th term of the AP 3 5 7 9 , , , ,......... 4 4 4 4 (v) the 15the term of the AP 40, 15,10,35,......... Sol: (i) The given A P is 9,13,17,21,.......... First term, 9 a Common difference, 13 9 4 d t h n term of the AP, 1 9 1 4 n a a n d n 20 t h term of the AP, 20 9 20 1 4 9 76 85 a (ii) The given AP is 20,17,14,11,.......... First term, 20 a Common difference, 17 20 3 d t h n term of the AP, 1 20 1 3 n a a n d n 35 t h term of the AP, 35 20 35 1 3 20 102 82 a (iii) The given AP is 2, 18, 50, 98,........... This can be re-written as 2,3 2,5 2,7 2,......... First term, 2 a Common difference, 3 2 2 2 2 d t h n term of the AP, 1 2 1 2 2 n a a n d n 18 t h term of the AP, 18 2 18 1 2 2 2 34 2 35 2 2450 a (iv) The given AP is 3 5 7 9 , , , ,......... 4 4 4 4 First term, 3 4 a Common difference, 5 3 2 1 4 4 4 2 d t h n term of the AP, 3 1 1 1 4 2 n a a n d n 9 t h t e r m of the AP, 9 3 1 3 19 9 1 4 4 2 4 4 a (v) The given AP is 40, 15,10,35,......... First term, 40 a Common difference, 15 40 25 d t h n term of the AP, 1 40 1 25 n a a n d n 15 t h term of the AP, 15 40 15 1 25 40 350 310 a 3. Find the 37 th term of the AP 3 1 1 6,7 ,9 ,11 ,......... 4 2 4 Sol: The given AP is 3 1 1 6,7 ,9 ,11 ,......... 4 2 4 First term, 6 a and common difference, 3 31 31 24 7 7 6 6 4 4 4 4 d Now, 37 37 1 36 T a d a d 7 6 36 6 63 69 4 37 t h term = 69 4. Find the 25 th term of the AP 1 1 5,4 , 4,3 ,3,........ 2 2 Sol: The given AP is 1 1 5,4 , 4,3 ,3,........ 2 2 First term = 5 Common difference 1 9 9 10 1 4 5 5 2 2 2 2 1 5 2 a a n d d Now, 25 25 1 24 T a d a d 1 5 24 5 12 7 2 25 t h term 7 5. Find the nth term of each of the following Aps: (ii) 16, 9, 2, - Sol: (i) (6n 1) (ii) (23 7n) 6. If the nth term of a progression is (4n 10) show that it is an AP. Find its (i) first term, (ii) common difference (iii) 16 the term. Sol: 4 10 n n T [Given] 1 4 1 10 6 T 2 3 4 4 2 10 2 4 3 10 2 4 4 10 6 T T T Clearly, 2 6 2 2 6 2 4 (Constant) So, the terms 6, 2,2,6,...... forms an AP. Thus we have (i) First term 6 (ii) Common difference = 4 (iii) 16 1 15 6 15 4 54 T a n d a d 7. Sol: In the given AP, 6 a and 10 6 4 d Suppose that there are n terms in the given AP. Then, 174 n T 1 174 a n d 6 1 4 174 2 4 174 4 172 43 n n n n Hence, there are 43 terms in the given AP. 8. How many terms are there in the AP 41, 3 Sol: In the given AP, 41 38 41 3 a a n d d Suppose that there are n terms in the given AP. Then 8 n T 1 8 41 1 3 8 44 3 8 3 36 12 a n d n n n n Hence, there are 12 terms in the given AP. 9. How many terms are there in the AP 1 18,15 ,13,....., 47. 2 ? Sol: The given AP is 1 18,15 ,13,....., 47. 2 First term, 18 a Common difference, 1 31 31 36 5 15 18 18 2 2 2 2 d Suppose there re n terms in the given AP. Then,Read More
|
124 videos|457 docs|77 tests
|
Related Exams
About this Document
|
4.96/5 Rating |
|
Nov 08, 2024 Last updated |
Document Description: RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation.
The notes and questions for RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) have been prepared according to the Class 10 exam syllabus. Information about RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) covers topics
like and RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) Example, for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Arithmetic Progressions (Exercise 5A).
Introduction of RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) in English is available as part of our Mathematics (Maths) Class 10
for Class 10 & RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) 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 5A) RD Sharma Solutions | Mathematics (Maths) Class 10
Description
Full syllabus notes, lecture & questions for Arithmetic Progressions (Exercise 5A) 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 5A)
In this doc you can find the meaning of RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) defined & explained in the simplest way possible. Besides explaining types of
RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) tests, examples and also practice Class 10
tests
|
Explore Courses for Class 10 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
study material
,Arithmetic Progressions (Exercise 5A) RD Sharma Solutions | Mathematics (Maths) Class 10
,Viva Questions
,Arithmetic Progressions (Exercise 5A) RD Sharma Solutions | Mathematics (Maths) Class 10
,Arithmetic Progressions (Exercise 5A) RD Sharma Solutions | Mathematics (Maths) Class 10
,Objective type Questions
,practice quizzes
,Summary
,shortcuts and tricks
,Important questions
,Previous Year Questions with Solutions
,ppt
,Free
,Extra Questions
,Semester Notes
,past year papers
,mock tests for examination
,MCQs
,video lectures
,Exam
,Sample Paper
;
Additional Information about RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) for Class 10 Preparation
RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) Free PDF Download
The RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) 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 5A) now and kickstart your journey towards success in the Class 10 exam.
Importance of RD Sharma Solutions: Arithmetic Progressions (Exercise 5A)
The importance of RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) 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 5A) Notes
RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) 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 5A).
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 5A) Notes on EduRev are your ultimate resource for success.
RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) Class 10 Questions
The "RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) 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 5A) on the App
Students of Class 10 can study RD Sharma Solutions: Arithmetic Progressions (Exercise 5A) 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 5A),
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 5A) is prepared as per the latest Class 10 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup