Comprehensive CBSE Question Bank in Mathematics X (Term-II)

CBSE II Question Bank in Mathematics STANDARD/BASIC CLASS 10 Features Short Answer Type Questions Long Answer Type Questions Strictly Based on the Latest CBSE Term-wise Syllabus Case Study Based MCQs Chapter-Wise Important Results and Formulae Very Short Answer Type Questions



Comprehensive CBSE Question Bank in Mathematics Standard/Basic Term–II (For Class X)



Comprehensive CBSE Question Bank in Mathematics Standard/Basic Term–II (For Class X) (According to the Latest CBSE Examination Pattern) By Dr. V.K. Soni M.Sc. (Maths), M.Ed., Ph.D Senior Lecturer Deptt. of Mathematics Lovely Professional University, Phagwara Formerly, Head Deptt. of Mathematics Army School, Amritsar Cantt. Punjab   laxmi Publications (P) Ltd (An iso 9001:2015 company) bengaluru • chennai • guwahati • hyderabad • jalandhar Kochi • kolkata • lucknow • mumbai • ranchi new delhi

Comprehensive CBSE Question Bank in Mathematics-Standard/Basic–X (Term-II) Copyright © by Laxmi Publications Pvt., Ltd. All rights reserved including those of translation into other languages. In accordance with the Copyright (Amendment) Act, 2012, no part of this publication may be reproduced, stored in a retrieval system, translated into any other language or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise. Any such act or scanning, uploading, and or electronic sharing of any part of this book without the permission of the publisher constitutes unlawful piracy and theft of the copyright holder’s intellectual property. If you would like to use material from the book (other than for review purposes), prior written permission must be obtained from the publishers. Printed and bound in India Typeset at : J.R. Enterprises, Delhi New Edition ISBN : 978-93-93268-83-9 Limits of Liability/Disclaimer of Warranty: The publisher and the author make no representation or warranties with respect to the accuracy or completeness of the contents of this work and specifically disclaim all warranties. The advice, strategies, and activities contained herein may not be suitable for every situation. In performing activities adult supervision must be sought. Likewise, common sense and care are essential to the conduct of any and all activities, whether described in this book or otherwise. Neither the publisher nor the author shall be liable or assumes any responsibility for any injuries or damages arising here from. The fact that an organization or Website if referred to in this work as a citation and/or a potential source of further information does not mean that the author or the publisher endorses the information the organization or Website may provide or recommendations it may make. Further, readers must be aware that the Internet Websites listed in this work may have changed or disappeared between when this work was written and when it is read. All trademarks, logos or any other mark such as Vibgyor, USP, Amanda, Golden Bells, Firewall Media, Mercury, Trinity, Laxmi appearing in this work are trademarks and intellectual property owned by or licensed to Laxmi Publications, its subsidiaries or affiliates. Notwithstanding this disclaimer, all other names and marks mentioned in this work are the trade names, trademarks or service marks of their respective owners. & Bengaluru 080-26 75 69 30 & Chennai 044-24 34 47 26 Branches & Guwahati 0361-254 36 69 & Hyderabad 040-27 55 53 83 & Jalandhar 0181-222 12 72 & Kochi 0484-405 13 03 & Kolkata 033-40 04 77 79 & Lucknow 0522-430 36 13 Published in India by & Ranchi 0651-224 24 64 Laxmi Publications (P) Ltd. C—00000/021/12 Printed at : Ajit Printing Press, Delhi. (An ISO 9001:2015 Company) 113, GOLDEN HOUSE, GURUDWARA ROAD, DARYAGANJ, NEW DELHI - 110002, INDIA Telephone : 91-11-4353 2500, 4353 2501 www.laxmipublications.com [email protected]

CONTENTS Chapters ... Pages 1. Quadratic Equations ... 1 2. Arithmetic Progressions ... 3. Circles ... 39 4. Constructions ... 77 5. Some Applications of Trigonometry (Heights and Distance) ... 104 6. Surface Areas and Volumes ... 112 7. Statistics 152 198

Course Structure Class–X (2021–22) Second Term NO. UNIT NAME MARKS I ALGEBRA (Cont.) 10 II GEOMETRY (Cont.) 09 III TRIGONOMETRY (Cont.) 07 IV MENSURATION (Cont.) 06 V STATISTICS & PROBABILITY (Cont.) 08 Total 40 INTERNAL INTERNAL ASSESSMENT 10 ASSESSMENT TOTAL 50 Periodic Tests MARKS TOTAL MARKS Multiple Assessments 03 10 marks for the term 02 Portfolio Student 02 Enrichment 03 Activities- practical work UNIT-I: ALGEBRA 1. QUADRATIC EQUATIONS (10) Periods Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). Solutions of quadratic equations (only real roots) by factorization, and by using quadratic formula. Relationship between discriminant and nature of roots. Situational problems based on quadratic equations related to day to day activities (problems on equations reducible to quadratic equations are excluded) 2. ARITHMETIC PROGRESSIONS Motivation for studying Arithmetic Progression Derivation of the nth term and sum of the first n terms of A.P. and their application in solving daily life problems. (Applications based on sum to n terms of an A.P. are excluded)

UNIT-II: GEOMETRY 3. CIRCLES Tangent to a circle at, point of contact 1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact. 2. (Prove) The lengths of tangents drawn from an external point to a circle are equal. 4. CONSTRUCTIONS 1. Division of a line segment in a given ratio (internally). 2. Tangents to a circle from a point outside it. UNIT-III: TRIGONOMETRY 5. SOME APPLICATIONS OF TRIGONOMETRY HEIGHTS AND DISTANCES–Angle of elevation, Angle of Depression. Simple problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation/depression should be only 30°, 45°, 60°. UNIT-IV: MENSURATION 6. SURFACE AREAS AND VOLUMES 1. Surface areas and volumes of combinations of any two of the following: cubes, cuboids, 2. spheres, hemispheres and right circular cylinders/cones. Problems involving converting one type of metallic solid into another and other mixed problems. (Problems with combination of not more than two different solids be taken). UNIT-V: STATISTICS & PROBABILITY 7. STATISTICS Mean, median and mode of grouped data (bimodal situation to be avoided). Mean by Direct Method and Assumed Mean Method only



2CHAPTER ARITHMETIC PROGRESSIONS IMPORTANT POINTS TO REMEMBER FOR QUICK REVISIONS In this chapter, we will study the following points to remember. ∑ Arithmetic Progression (A.P.): An arithmetic progression (A.P.) is an ordered list of numbers in which each term progresses (i.e., increases or decreases) successively by a constant/fixed number called, the common difference (d). Each term of an A.P., except the first term, is obtained by adding the common difference (d) to the preceding term. ∑ The General Form of an A.P.: The general form of an A.P. with the first term, a, and common difference, d is given by: a, a + d, a + 2d, a + 3d, ............... ∑ The General (or nth) Term of an A.P.: In an A.P. with first term a and common difference d, the General term (or nth term) is given by Tn = a + (n – 1) d. ∑ Selection of Terms in an A.P.: (i) For three terms, take: a – d, a and a + d (ii) For four terms, take: a – 3cd, a – d, a + d and a + 3d (iii) For five terms, take: a – 2d, a – d, a, a + d and a + 2d and so on. ∑ The Sum Formula: The sum of the first, n terms of an A.P. is given by: Sn = n [2a + (n – 1)d] ... 1st Form 2 Sn = n (a + l) ... 2nd Form 2 where l is the last term in Sn i.e., l = Tn = a + (n – 1) d In the above Sum Formula, use the 1st Form when the common difference, d is known and use the 2nd Form when the last term, l is known. ∑ nth Term in Terms of Sn : If the sum, Sn of n terms of an A.P., is given, then the nth term of the A.P. can be computed by using the formula Tn = Sn – Sn–1 , for n > 1. 39

40 MATHEMATICS-X PART A VERY SHORT ANSWER TYPE QUESTIONS Illustrative Examples EXAMPLE 1. (i) If 4 , a, 2, are three consecutive terms of an A.P, then find the value of a. 5 (ii) If the numbers n – 2, 4n – 1 and 5n + 2 are in A.P, find the value of n. [NCERT (EP)] SOLUTION. (i) Here, we have 4 , a, 2 as three consecutive terms of an A.P. 5 fi a– 4 = 2 – a By Def. of an A.P. 5 fi a + a = 2 + 4 5 fi 2a = 10 + 4 = 14 fi a = 14 = 7 55 2¥5 5 Hence, the required value of a is 7 . By Def. of an A.P. 5 (ii) As n – 2, 4n – 1, 5n + 2 are in A.P fi (4n – 1) – (n – 2) = (5n + 2) – (4n – 1) fi 3n + 1 = n + 3 fi 3n – n = 3 – 1 fi 2n = 2 fi n= 2 =1 Hence, the required value of n is 1. 2 EXAMPLE 2. Write first four terms of the A.P, when the first term a and common difference d are as follows: (i) a = 10, d = 10 (ii) a = – 2, d = 0 (iii) a = 4, d = - 3 (iv) a = - 1, d= 1 (v) a = - 1.25, d = – 0.25 [NCERT] 2 SOLUTION. The first four terms of an A.P, whose first term is a and the common difference is d are a, a + d, a + 2d, a + 3d ...(1) Using the General form of an A.P. (i) Here, a = 10, d = 10 Putting a = 10, d = 10 in (1), we get the first four terms of an A.P, as 10, 10 + 10, 10 + 2 (10), 10 + 3 (10). i.e., 10, 20, 30, 40

ARITHMETIC PROGRESSIONS 41 (ii) Here, a = – 2, d = 0 Putting a = – 2, d = 0 in (1), we get the first four terms of an A.P, as – 2, – 2 + 0, – 2 + 2 (0), – 2 + 3 (0). i.e., – 2, – 2, – 2, – 2 (iii) Here, a = 4, d = – 3 Putting a = 4, d = – 3 in (1), we get the first four terms of an A.P, as 4, 4 + (– 3), 4 + 2 (– 3), 4 + 3 (– 3) i.e., 4, 1, – 2, – 5 (iv) Here, a = –1, d= 1 2 Putting a = –1, d= 1 in (1), we get the first four terms of an A.P, as 2 –1, –1 + 1 , –1 + 2 ÁÊË 1 ˜ˆ¯ , –1 + 3 ÁÊË 1 ¯˜ˆ i.e., –1, – 1 , 0, 1 2 2 2 2 2 (v) Here, a = –1.25, d = –0.25 Putting a = –1.25, d = –0.25 in (1), we get the first four terms of an A.P, as –1.25, –1.25 + (–0.25), –1.25 + 2 (–0.25), –1.25 + 3 (–0.25) i.e., –1.25, –1.50, –1.75, –2.0 EXAMPLE 3. In an A.P, if the common difference (d) = –4 and the seventh term ( a7) is 4, find the first term. [CBSE 2018] SOLUTION. Common difference of A.P (d) = –4 and 7th term of A.P. (a7) = 4 i.e., a7 = a + (7 – 1)d   an = a + (n – 1)d fi 4 = a + 6 ¥ (– 4) fi a = 28 \\ First term = 28. EXAMPLE 4. For the following A.Ps, write the first term and the common difference: (i) 3, 1, -1, -3 ............... (ii) -5, -1, 3, 7, ............... (iii) 1 , 5 , 9 , 13 , ............... (iv) 0.6, 1.7, 2.8, 3.9, ............... [NCERT] 3 3 3 3 SOLUTION. (i) Here, the first term of the A.P, a = 3 The common difference, d = T2 - T1 = 1 - 3 = -2 Hence, the first term of the A.P is 3 and the common difference is -2. (ii) Here, the first term of the A.P, a = –5 The common difference, d = T2 - T1 = -1 - (-5) = -1 + 5 = 4 Hence, the first term of the A.P is -5 and the common difference is 4. (iii) Here, the first term of the A.P, a= 1 3

42 MATHEMATICS-X The common difference, d = T2 - T1 = 5 – 1 = 5-1 = 4 3 3 3 3 Hence, the first term of the A.P, is 1 and the common difference is 4 . 3 3 (iv) Here, the first term of the A.P, a = 0.6 The common difference, d = T2 - T1 = 1.7 - 0.6 = 1.1 Hence, the first term of the A.P is 0.6 and the common difference is 1.1. EXAMPLE 5. Which term of the A.P: 3, 8, 13, 18, ..............., is 78? [NCERT, CBSE 2012] SOLUTION. The given A.P is 3, 8, 13, 18, ............... Here, we have The first term of the A.P, a = 3 The common difference of the A.P, d = 8 - 3 = 5 The term-number, n=? Required nth term of the A.P, Tn = 78 Using the General term formula, we have Tn = a + (n - 1) d fi 78 = 3 + (n - 1) 5 fi 78 = 5n - 2 fi 78 = 3 + 5n - 5 fi 5n = 80 fi n= 80 = 16 5 So, 16th term of the A.P is 78. EXAMPLE 6. Find the number of terms in each of the following A.Ps: (i) 7, 13, 19, ..............., 205 (ii) 18, 15 1 , 13, ..............., - 47 2 [NCERT, CBSE 2012, 19] SOLUTION. (i) The given A.P is 7, 13, 19, ..............., 205 Here, we have The first term of the A.P, a = 7 The common difference of the A.P, d = 13 - 7 = 6 Suppose, the number of term in the A.P = n Required Then, nth term of the A.P, Tn = 205 Using the General term formula, we have Tn = a + (n - 1) d fi 205 = 7 + (n - 1) 6 fi 205 = 1 + 6n fi 205 = 7 + 6n - 6 fi 6n = 205 - 1 = 204 fi n= 204 = 34 6 Hence, there are 34 terms in the given A.P.

ARITHMETIC PROGRESSIONS 43 (ii) The given A.P is 18, 15 1 , 13, ..............., -47 2 Here, we have The first term of the A.P, a = 18 The common difference of the A.P, d= ÊËÁ15 1 ˆ - 18 = 31 - 36 = - 5 2 ¯˜ 2 2 Suppose, the number of term in the A.P = n Required Then, nth term of the A.P, Tn = - 47 Using the General term formula, we have Tn = a + (n - 1) d -47 = 18 + (n - 1) Ê - 5ˆ ËÁ 2 ˜¯ fi -47 = 18 - 5n + 5 fi -135 = - 5n 2 2 fi 2 (-47) = 36 - 5n + 5 fi 5n = 135 fi -94 = 41 - 5n fi n= 135 = 27 fi -41 - 94 = - 5n 5 Hence, there are 27 terms in the given A.P. EXAMPLE 7. Find the 20th term from the end of the A.P 3, 8, 13, ............... , 253. [NCERT, CBSE 2012, 19] SOLUTION. Here, the given A.Ps is 3, 8, 13, ..............., 253 Last term, l = 253 and C.D. ‘d’ = 8 – 3 = 5 \\ T–20 = [253 – (20 – 1)5] = 253 – 95 = 158. Hence, the 20th term from the last is 158. EXAMPLE 8. Find the sum of the following A.Ps: 1 , 1 , 1 , ............... to 11 terms [NCERT] 15 12 10 SOLUTION. Here, the given A.P is 1 , 1 , 1 , ............... to 11 terms. 15 12 10 Then, a = 1 , n = 11 and 15 d= 1 - 1 = 15 - 12 = 3 = 1 12 15 12 ¥ 15 12 ¥ 15 60 Now, Sn = n [2a + (n - 1)d ] Using the Sum Formula 2

44 MATHEMATICS-X i.e., S11 = 11 ÍÈÎ2 ¥ ËÁÊ 1 ¯ˆ˜ + (11 - 1) 1 ˘ = 11 È2 + (10) 1 ˘ 2 15 60 ˚˙ 2 ÍÎ 15 60 ˚˙ = 11 È2 + 1˘ = 11 È 4 + 5 ˘ = 11 È 9 ˘ = 11 È 3 ˘ = 33 2 ÍÎ 15 6 ˙˚ 2 ÎÍ 30 ˚˙ 2 ÎÍ 30 ˚˙ 2 ÍÎ10 ˚˙ 20 So, the sum of the first 11 terms of the given A.P is 33 . 20 EXAMPLE 9. (i) Find the sum of first five multiples of 3. [NCERT (EP)] (ii) Find the sum of first seven numbers which are multiples of 2 as well as 9? [NCERT (EP)] SOLUTION. (i) Here, the first five multiples of 3 are: 3, 6, 9, 12 and 15. Then, a = 3, l = 15 and n = 5 \\ S5 = 5 [3 + 15] = 5 ¥ 18 = 45.  Sn = n (a + l) 2 2 2 Hence, the required sum of first 5 multiples of 3 is 45. (ii) To find seven numbers which are multiples of 2 as well as 9, we take LCM of 2 and 9. LCM of 2 and 9 = 18 \\ The first 7 multiples of 2 and 9 are: 18, 36, 54, ............... Then a = 18, d = 36 – 18 = 18 \\ S7 = 7 [2(18) + (7 – 1)18]  Sn = n [2a + (n – 1)d 2 2 = 7 [2 ¥ 18 + 6 ¥ 18] = 7 ¥ 18 [2 + 6] 22 = 7 ¥ 9 ¥ 8 = 504 Hence, the required sum of 1st 7 multiples of both 2 and 9 is 504. TEST YOUR KNOWLEDGE 1. If k + 1, 3k and 4k + 2 be any three consecutive terms of an A.P., find the value of k. [CBSE 2012, 14] 2. (i) Which term of the A.P: 21, 18, 15, ............... is zero? [CBSE 2012] (ii) Which term of the A.P: 21, 18, 15, ............... is -81? [NCERT, CBSE 2012] (iii) Check whether 301 is a term of the list of numbers 5, 11, 17, 23, ...............? [NCERT] 3. (i) Find the 10th term from end of the A.P: 4, 9, 14, ..............., 254. [CBSE 2012] (ii) Find the 6th term from end of the A.P: 5, 2, –1, –4, ..............., –31. 4. (i) Find the sum of the first 22 terms of the A.P: 8, 3, -2, ............... [CBSE 2012] [NCERT] (ii) Find the sum of first five positive integers divisible by 6. [CBSE 2012] (iii) If the first term of an A.P is –5 and the common difference is 2, then find the sum of the first 6 terms. [CBSE 2012]

ARITHMETIC PROGRESSIONS 45 5. (i) Find the sum of first n natural numbers. [NCERT] (ii) Find the sum of first 20 odd natural numbers. [CBSE 2012] (iii) Find the sum of first 1000 positive integers. [NCERT] 6. (i) How many two-digit numbers are divisible by 3? [CBSE 2012] (ii) Is series 3 , 6 , 9 , 12 , ........................... an A.P.? Give reason. [CBSE 2015] 7. Which term of the A.P. 27, 24, 21, ... is zero? [CBSE 2020] 8. In an Arithmetic progression, if d = –4, n = 7, an = 4, then find a. [SQP 2020] 9. If 3 , a, 4 are three consecutive terms of of an A.P., then find the value of a. 5 [Comp. 2021] 10. In an A.P., if the common difference d = –3 and the eleventh term a11 = 15, then find the first term. [Comp. 2021] PART B SHORT ANSWER TYPE-I QUESTIONS Illustrative Examples EXAMPLE 1. Which of the following are A.Ps? If they form an A.P, find the common difference d and write three more terms: (i) 3, 3 + 2 , 3 + 2 2 , 3 + 3 2 , ............... (ii) 0, - 4, - 8, - 12, ............... (iii) - 1 , - 1 , - 1 , – 1 , ............... 2 2 2 2 (iv) 3 , 6 , 9 , 12 , ............... (v) 12, 32, 52, 72, ............... (vi) 12, 52, 72, 73, ............... [NCERT] SOLUTION. (i) Here, the given list of numbers is 3, 3 + 2 , 3 + 2 2 , 3 + 3 2 , ............... T2 - T1 = (3 + 2 ) - 3 = 2 T3 - T2 = (3 + 2 2 ) - (3 + 2 ) = 2 (2 - 1) = 2  T3 - T2 = T2 - T1 d= 2 Hence, the given list of numbers forms an A.P. \\ Next three terms after the last given term are 3 + 3 2 + 2 = 3 + 4 2; 3 + 4 2 + 2 = 3 + 5 2; 3+5 2 + 2 =3+6 2 Hence, the next three terms are 3 + 4 2 , 3 + 5 2 , 3 + 6 2 .

46 MATHEMATICS-X (ii) Here, the given list of numbers is 0, -4, - 8, -12, ............... T2 - T1 = - 4 - 0 = - 4 T3 - T2 = - 8 - (- 4) = - 8 + 4 = - 4 T4 - T3 = - 12 - (- 8) = - 12 + 8 = - 4 i.e., Tn + 1 - Tn is the same every time. Hence, the given list of numbers forms an A.P. So, d = - 4 \\ Next three terms after the last given term are - 12 + (- 4) = - 12 - 4 = -16; - 16 + (- 4) = - 16 - 4 = -20; - 20 + (- 4) = - 20 - 4 = -24 Hence, the next three terms are -16, -20, -24. (iii) Here, the given list of numbers is - 1 , - 1 , - 1 , - 1 , ............... 2 2 2 2 T2 - T1 = - 1 - ÁÊË - 1 ˜ˆ¯ = 0; 2 2 T3 - T2 = - 1 - ÁÊË - 1 ˜ˆ¯ = 0; 2 2 T4 - T3 = - 1 - Ê - 1ˆ =0 2 ÁË 2 ˜¯ i.e., Tn + 1 - Tn is the same every time. Hence, the given list of numbers forms an A.P. So, d = 0 \\ Next three terms after the last given term are - 1 + 0 = – 1 , - 1 , - 1 2 2 2 2 Hence, the next three terms are - 1 , - 1 , - 1 2 2 2 (iv) Here, the given list of numbers is 3 , 6 , 9 , 12 , ............... ( ) ( )T2 - T1 = 6 - 3 = 3 ¥ 2 - 3 = 3 2 - 1 T3 - T2 = 9 - 6 = (3) - ( 6 ) = 3 - 6 T3 - T2 π T2 - T1 fi The given list of numbers does not form an A.P. (v) Here, the given list of numbers is 12, 32, 52, 72, ............... T2 - T1 = 32 - 12 = 9 - 1 = 8 T3 - T2 = 52 - 32 = 25 - 9 = 16 T3 - T2 π T2 - T1 fi The given list of numbers does not form an A.P.

ARITHMETIC PROGRESSIONS 47 (vi) Here, the given list of numbers is 12, 52, 72, 73, ............... T2 - T1 = 52 - 12 = 52 - 1 = 24 T3 - T2 = 72 - 52 = 49 - 25 = 24 T4 - T3 = 73 - 72 = 73 - 49 = 24 i.e., Tn + 1 - Tn is the same every time. Hence, the given list of numbers forms an A.P. So, d = 24 Next three terms after the last given term are 73 + 24 = 97; 97 + 24 = 121; and 121 + 24 = 145 Hence, the next three terms are 97, 121, 145. EXAMPLE 2. (i) How many three digit numbers are divisible by 7? [NCERT] (ii) Find the number of all three-digit natural numbers which are divisible by 9. [CBSE 2013] SOLUTION. (i) The first three digit number divisible by 7 = 105 The last three digit number divisible by 7 = 994 Thus, we have to find the number of terms in the list of numbers given below: 105, 112, 119, ....., 994 ...(1) Now, T2 - T1 = 112 - 105 = 7; T3 - T2 = 119 - 112 = 7  T3 - T2 = T2 - T1 = 7 fi The list of numbers in (1) forms an A.P. 1st term of an A.P, a = 105; C.D of the A.P, d = 7 Let the number of terms in the A.P be n Then, last term of the A.P, l = 994 Using the General term formula of an A.P, we get a + (n - 1)d = 994 fi 105 + (n - 1)7 = 994 fi (n - 1)7 = 994 - 105 = 889 fi n-1 = 889 = 127 \\ 7 n = 128 Hence, there are 128 numbers of three digit numbers which are divisible by 7. (ii) Here, the 3-digit natural numbers which are divisible by 9 are: 108, 117, 126, .............., 999 ...(1) Now, T2 – T1 = 117 – 108 = 9; T3 – T2 = 126 – 117 = 9  T2 – T1 = T3 – T2 = 9 fi The list of numbers in (1) form an A.P, with a = 108, d = 9, l = 999

48 MATHEMATICS-X Let the number of terms in A.P (1) be n. Then, 999 = 108 + (n – 1)9   l = a + (n – 1) 9 fi 9 (n – 1) = 999 – 108 fi 9n = 891 + 9 = 900 fi 9n – 9 = 891 fi n= 900 = 100 9 Hence, there are 100 numbers of 3-digit natural numbers which are divisibly by 9. EXAMPLE 3. How many multiples of 4 lie between 10 and 250? [NCERT, CBSE 2019] SOLUTION. Here, multiples of 4 between 10 and 250 are: 12, 16, 20, ..............., 248 which is an A.P with common difference 4. First multiple of 4 between 10 and 250 = 12 and last multiple of 4 between 10 and 250 = 248 Let the nth term be 248 Then, first term of the A.P, a = 12 Common difference of the A.P, d = 4 = (16 - 12) and nth term of the A.P, Tn = 248 Using the General term formula of an A.P, we have Tn = a + (n - 1)d fi a + (n - 1) d = 248   Tn = 248 fi 4n + 8 = 248 fi 12 + (n - 1)4 = 248 fi 4n = 248 - 8 = 240 fi 12 + 4n - 4 = 248 fi n= 240 = 60 4 Hence, there are 60 multiples of 4 between 10 and 250. EXAMPLE 4. The 17th term of an A.P exceeds its 10th term by 7. Find the common difference. [NCERT] SOLUTION. Suppose, the first term of the A.P = a The common difference of the A.P = d Using the General term formula, we have Tn = a + (n - 1) d ...(1) fi T17 = a + (17 - 1) d = a + 16d ...(2) and T10 = a + (10 - 1) d = a + 9d According to the problem, T17 - T10 = 7 Using (1) and (2) fi (a + 16d) - (a + 9d) = 7 fi 16d - 9d = 7 fi 7d = 7 fi d=1 Hence, the required common difference is 1. EXAMPLE 5. Find the middle term(s) in the A.P. 20, 16, 12, ..............., (–176) [CBSE 2012] ...(1) SOLUTION. Here, the given A.P is 20, 16, 12, ..............., (–176) Then, a = 20, l = –176, and d = 16 – 20 = –4.

54 MATHEMATICS-X EXAMPLE 2. Find the 31st term of an A.P whose 11th term is 38 and the 16th term is 73. [NCERT, CBSE 2012] SOLUTION. Suppose, the first term of the A.P, = a The common difference of the A.P, = d 31st term of the A.P, T31 = ? Required 11th term of the A.P, T11 = 38 Given 16th term of the A.P, T16 = 73 Given Using the nth term formula, we have Tn = a + (n - 1) d fi T11 = a + (11 - 1) d fi 38 = a + 10d fi a + 10d = 38 ...(1) fi T16 = a + (16 - 1) d fi 73 = a + 15d fi a + 15d = 73 ...(2) Now, we have to solve the simultaneous equations (1) and (2), subtracting (1) from (2), a + 15d = + 73 ...(2) + a + 10d = + 38 ...(1) –– – 5d = 35 fi d= 35 =7 ...(3) 5 Putting the value of d from (3) in (1), we get a + 10 (7) = 38 fi a + 70 = 38 fi a = -70 + 38 = -32 ...(4) Now, putting the values of a = -32, d = 7 and n = 31 in the General term formula, we get T31 = -32 + (31 - 1)(7) = -32 + 30(7) = -32 + 210 = 178 Hence, the required 31st term of the A.P is 178. EXAMPLE 3. If the 3rd and 9th terms of an A.P are 4 and -8 respectively, which term of this A.P is zero? [NCERT] SOLUTION. Here, we have The third term of the A.P, T3 = 4 The 9th term of the A.P, T9 = -8 Let ‘a’ be the first term of the A.P and ‘d’ be the common difference of the A.P. Using the General term formula, we get Tn = a + (n - 1)d fi T3 = a + (3 - 1)d Here, the term number n = 3 fi 4 = a + 2d fi a + 2d = 4 ...(1) T9 = a + (9 - 1)d Here, the term number n = 9 fi -8 = a + 8d fi a + 8d = -8 ...(2) Now, we have to solve the simultaneous equations (1) and (2).

ARITHMETIC PROGRESSIONS 55 Subtracting (1) from (2), ...(2) a + 8d = -8 ...(1) a + 2d = + 4 –– + 6d = –12 fi 6d = -12 fi d = - 12 = -2 ...(3) 6 Putting the value of d from (3) in (1), we get a + 2(- 2) = 4 fi a-4=4 fi a=4+4=8 ...(4) Now, putting the values of a = 8, d = -2 and Tn = 0, in the General term formula, we get Tn = a + (n - 1) d fi 0 = 8 + (n - 1) (-2) Since, it is given that Tn = 0 fi 0 = 8 + (-2n + 2) = 10 - 2n fi 2n = 10 fi n= 10 =5 2 Hence, the 5th term of the A.P is zero. EXAMPLE 4. Two A.Ps have the same common difference. The difference between their 100th terms are 100, what is the difference between their 1000th terms? [NCERT] SOLUTION. Let first terms of two A.Ps be a and A respectively Let there common difference be d. Then, using General term formula, we have Tn = A + (n - 1)d ...(1) fi T100 = A + (100 - 1)d fi t100 = a + (100 - 1)d ...(2) fi T100 = A + 99d ...(3) fi t100 = a + 99d Case (I): According to the problem, T100 - t100 = A + 99d - a - 99d ...(4) fi 100 = A - a fi A - a = 100 Case (II): Again, using General term formula, we have T1000 = A + (1000 - 1)d fi t1000 = a + (1000 - 1)d fi T1000 = A + 999d fi t1000 = a + 999d According to the problem, T1000 - t1000 = A + 999d - a - 999d = A - a = 100 By using (4) Hence, the required difference between the 1000th term of the two A.Ps is 100.

56 MATHEMATICS-X EXAMPLE 5. The sum of the 4th and 8th terms of an A.P is 24 and the sum of the 6th and 10th term is 44. Find the first three terms of the A.P. [NCERT, CBSE 2012] SOLUTION. Suppose, the 1st term of the A.P = a and, the common difference of the A.P = d Using the General term formula, Tn = a + (n - 1)d, we have T4 = a + (4 - 1)d = a + 3d ...(1) ...(2) T8 = a + (8 - 1)d = a + 7d ...(3) ...(4) T6 = a + (6 - 1)d = a + 5d T10 = a + (10 - 1)d = a + 9d According to the problem, Case (I): T4 + T8 = 24 (a + 3d) + (a + 7d) = 24 fi 2a + 10d = 24 Using (1) and (2) ...(5) fi 2(a + 5d) = 24 fi a + 5d = 24 = 12 2 Using (3) and (4) Case (II): T6 + T10 = 44 ...(6) (a + 5d) + (a + 9d) = 44 fi 2a + 14d = 44 fi 2(a + 7d) = 44 fi a + 7d = 44 = 22 2 Now, we have to solve simultaneous equations (5) and (6). So, subtracting (6) from (5), we get -2d = -10 fi d= -10 =5 -2 Putting the value of d in (5), we get a + 5d = 12 fi a + 5(5) = 12 fi a = 12 - 25 = -13 The 1st term of the A.P, a = -13 The 2nd term of the A.P The 3rd term of the A.P = a + d = -13 + 5 = -8 = a + 2d = -13 + 2(5) = -13 + 10 = -3 Hence, the required first three terms of the A.P are -13, -8, -3. EXAMPLE 6. The sum of the 2nd and the 7th terms of an A.P is 30. If its 15th term is 1 less than twice its 8th term, find the A.P. [CBSE 2014] SOLUTION. Let a be the first term and d, the common difference of the A.P. Using the General term formula, Tn = a + (n - 1)d, we have

ARITHMETIC PROGRESSIONS 57 T2 = a + (2 - 1)d = a + d ...(1) T7 = a + (7 - 1)d = a + 6d ...(2) T8 = a + (8 - 1)d = a + 7d ...(3) T15 = a + (15 - 1)d = a + 14d ...(4) According to the problem, Case (I): T2 + T7 = 30 fi (a + d) + (a + 6d) = 30 Using (1) and (2) ...(5) fi 2a + 7d = 30 Case (II): T15 = 2T8 – 1 Using (3) and (4) fi a + 14d = 2(a + 7d) – 1 fi a + 14d = 2a + 14d – 1 a=1 ...(6) Putting the value of a from (6) in (5), we get 2(1) + 7d = 30 fi 7d = 30 – 2 = 28 fi d= 28 =4 7 Hence, the required A.P is 1, 5, 9, 13, ............... EXAMPLE 7. If the seventh term of an A.P is 1 and its ninth term is 1 , find its 63rd term. 9 7 [CBSE 2014] SOLUTION. Let a be the first term and d, the common difference of the A.P. Using the General term formula, Tn = a + (n - 1)d, we have T7 = a + (7 - 1)d = a + 6d = 1 ...(1) 9 T9 = a + (9 - 1)d = a + 8d = 1 ...(2) 7 Subtracting (1) from (2), we get 8d – 6d = 1 – 1 = 9–7 2 7 9 63 = 63 fi 2d = 2 fi d= 1 ...(3) 63 63 Putting the value of d from (3) in (1), we get a + 6ÁËÊ 613˜¯ˆ = 1 9 fi a = 1 – 6 = 7–6 = 1 fi a= 1 ...(4) 9 63 63 63 63

58 MATHEMATICS-X \\ T63 = a + (63 – 1)d = 1 + (62) ¥ Ê 1ˆ Using (3) and (4) 63 ÁË 63˜¯ fi T63 = 1 + 62 fi T63 = 63 =1 63 63 63 Hence, the required 63rd term of the A.P is 1. Example 5.21 can be generalised as “if the pth term of an A.P. be 1 and qth q term be 1 , then its (pq)th term is 1.” p EXAMPLE 8. If m times the mth term of an A.P is equal to n times its nth term, show that the (m + n)th term of the A.P, is zero. [CBSE 2019] Or If m times the mth term of an A.P is equal to n times the nth term, find (m + n )th term of A.P. SOLUTION. Let a be the first term and d, the common difference of the A.P. According to the problem, mTm = nTn Using the General term formula of an A.P. fi m[a + (m - 1)d] = n[a + (n - 1)d] fi ma + (m2 - m)d = na +(n2 - n)d fi ma - na + (m2 - m)d - (n2 - n)d = 0 fi (m - n)a + (m2 - m - n2 + n)d = 0 fi (m - n)a + (m2 - n2 - m + n)d = 0 fi (m - n)a + [(m + n)(m - n) - (m - n)]d = 0 fi (m - n)[a + (m + n - 1)d] = 0 But m - n π 0  m π n \\ a + (m + n - 1)d = 0 fi Tm + n = 0 Hence, proved. EXAMPLE 9. If pth term of an A.P is q and qth term is p, then show that its nth term is (p + q - n). [CBSE 2015] SOLUTION. Let a be the first term and d be the common difference of the given A.P. Then, pth term of the A.P, Tp = a + (p - 1)d ...(1) qth term of the A.P, Tq = a + (q - 1)d ...(2) Now, according to the problem, we have Tp = q and Tq = p Given a + (p - 1)d = q ...(3) Using (1)

ARITHMETIC PROGRESSIONS 59 a + (q - 1)d = p ...(4) Using (2) Subtracting equation (4) from (3), we get (p - q)d = (q - p) fi d= – (p - q) = -1 p - q Putting d = -1 in (3), we get a + (p - 1)(–1) = q fi a = (p + q - 1) Using General term formula of an A.P, Tn = a + (n - 1)d = (p + q - 1) + (n - 1) (-1) = (p + q - n)   a = (p + q - 1) and d = -1 Hence, the nth term of the given A.P is (p + q - n) EXAMPLE 10. If the pth, qth, rth terms of an A.P be a, b, c respectively, then show that: a(q - r) + b(r - p) + c(p - q) = 0. SOLUTION. Let A be the first term and D be the common difference of an A.P. Then, pth term of the A.P, Tp = a qth term of the A.P, Tq = b rth term of the A.P, Tr = c Using the General term formula of an A.P, we get fi Tn = A + (n - 1) D ...(1) fi Tp = A + (p - 1) D fi Tq = A + (q - 1) D Tr = A + (r - 1) D \\ A + (p - 1) D = a A + (q - 1) D = b ...(2) A + (r - 1) D = c ...(3) Multiplying (1) by (q - r), (2) by (r - p) and (3) by (p - q) and adding, we get a(q - r) + b(r - p) + c(p - q) = A[(q - r) + (r - p) + (p - q)] + D[(p - 1)(q - r) + (q - 1) (r - p) + (r - 1)(p - q)] = (A ¥ 0) + D[pq – pr – q + r + qr – qp – r + p + rp – rq – p + q] = (A ¥ 0) + (D ¥ 0) = 0 \\ a(q - r) + b(r - p) + c(p - q) = 0 Hence proved. EXAMPLE 11. Find four numbers in an A.P whose sum is 20 and the sum of whose squares is 120. SOLUTION. Let the required numbers be (a - 3d), (a - d), (a + d), and (a + 3d). Sum of these numbers,

60 MATHEMATICS-X S = (a - 3d) + (a - d) + (a + d) + (a + 3d) = 4a ...(1) Sum of the squares of these numbers = (a - 3d)2 + (a - d)2 + (a + d)2+ (a + 3d)2 ...(2) = 4(a2 + 5d2) According to the problem, S = 20 Given fi 4a = 20 Using (1) fi a=5 ...(3) Also, 4(a2 + 5d2) = 120 Using (2) fi a2 + 5d2 = 30 fi 25 + 5d2 = 30  a = 5 fi 5d2 = 5 fi d = ±1 Hence, a = 5 and d = ±1. Case (I): When d = 1. The numbers are [5 - 3 (1)], (5 - 1), (5 + 1), and [5 + 3(1)] i.e., 2, 4, 6, 8. Case (II): When d = -1. The numbers are [5 - 3 (-1)], [5 - (-1)], [5 + (-1)], and [5 + 3(-1)] i.e., 8, 6, 4, 2. Hence, the required numbers are (2, 4, 6, 8) or (8, 6, 4, 2.) EXAMPLE 12. Find the sums of the following: [NCERT] - 5 + (- 8) + (- 11) + ............... + (- 230) SOLUTION. Here, the given A.P is - 5 + (-8) + (-11) + ............... + (-230). With a = –5, l = –230 and d = – 8 – (–5) = – 3 Let the total number of terms, be ‘n’ of the A.P ln = a + (n - 1)d Using the General term formula fi - 230 = - 5 + (n - 1) (- 3) fi 3n = 230 - 2 fi - 230 = - 5 - 3n + 3 fi 3n = 228 fi - 230 = - 2 - 3n fi n= 228 = 76 Now, sum of the first seventy-six terms, 3 S76 = ? Required n Using the Sum formula Now, Sn = 2 (a + l) fi S76 = 76 (- 5 - 230) = 38 (- 235) = - 8930 2 So, the required sum of the given A.P is - 8930. EXAMPLE 13. In an A.P. [NCERT] (i) given a = 3, n = 8, S = 192, find d. (ii) given l = 28, S = 144, and there are total 9 terms. Find a.

ARITHMETIC PROGRESSIONS 61 SOLUTION. (i) Here, we have a = 3, n = 8, Sn = 192 Common difference of the A.P, d = ? Required Using the General term formula of an A.P, we get Sn = n [2a + (n - 1)d ] 2 fi 192 = 8 [2(3) + (8 - 1)d] = 4[6 + 7d] 2 fi 6 + 7d = 192 = 48 fi 7d = 48 - 6 = 42 fi d=6 4 Hence, the required value of the common difference, d is 6. (ii) Here, we have l = 28, S = 144, n = 9 Required 1st term of the A.P, a = ? Using the Sum formula of an A.P, we have S= n [a + l] 2 fi 144 = 9 [a + 28] fi 288 = 9a + 252 2 fi 9a = 288 - 252 = 36 fi a= 36 =4 9 Hence, the required value of the 1st term, a is 4. EXAMPLE 14. (i) How many terms of the A.P: 3, 5, 7, 9, ............... must be added to get the sum 120? [CBSE 2020] (ii) How many terms of the A.P: 9, 17, 25, ............... must be taken to give a sum of 636? [NCERT] SOLUTION. (i) Here, the given A.P is 3, 5, 7, 9, ............... Then, a = 3, d = (5 - 3) = 2 Let the first nth term of the given A.P be added to get the sum 120. Then, Sn = 120 Using the Sum formula of A.P, we get n [2a + (n - 1)d] = 120 2 fi n [2 ¥ 3 + (n - 1)2 ] = 120 2 fi n [n + 2] = 120 fi n2 + 2n - 120 = 0 fi n2 + 12n - 10n - 120 = 0 fi (n + 12)(n - 10) = 0

62 MATHEMATICS-X fi Either n + 12 = 0 or n - 10 = 0 fi n = -12 fi n = 10 Rejecting n = -12,  n cannot be negative. Hence, the required number of terms in the A.P is 10. (ii) Here, the given A.P is 9, 17, 25 ............... With a = 9, d = (17 - 9) = 8 and Sn = 636 Common difference of the A.P Let the first n terms of the given A.P be added to get the sum 636. Using the Sum formula of A.P, we get n [2(a) + (n - 1)d] = 636 fi n [2 ¥ 9 + (n - 1) 8 ] = 636 2 2 fi n [9 + 4n - 4] = 636 fi 4n2 + 5n - 636 = 0 fi 4n2 + 53n - 48n - 636 = 0 Splitting the middle term fi 4n2 – 48n + 53n - 636 = 0 fi 4n(n - 12) + 53(n - 12) = 0 fi (n - 12) (4n + 53) = 0 By factorisation fi Either n - 12 = 0 or 4n + 53 = 0 fi n = 12 fi n= - 53 4 Rejecting n = - 53 ,  n cannot be negative and fraction 4 Hence, the required number of terms in the A.P is 12. EXAMPLE 15. Find the sum of odd numbers between 0 and 50. [NCERT, CBSE 2019] SOLUTION. The first odd number between 0 to 50 is 1 The last odd number between 0 and 50 is 49 \\ Odd numbers between 0 and 50 are 1, 3, 5, ..............., 49 ...(1) T2 - T1 = 3 - 1 = 2 T3 - T2 = 5 - 3 = 2 fi T3 - T2 = T2 - T1 = 2 fi List of odd numbers between 0 and 50 in (1) forms an A.P with a = 1, d = 2, l = 49 Let the number of terms of the A.P (1) be n. Using the General term formula of an A.P, we get a + (n - 1)d = Tn   Tn = l = 49 (Given) fi a + (n - 1)d = 49 fi 1 + (n - 1)2 = 49 fi 2n - 1 = 49 fi 2n = 50 fi n= 50 = 25. 2 So, we need to find the sum of 25 terms in the A.P (1).

ARITHMETIC PROGRESSIONS 63  Sn = n [a + l] fi S25 = 25 [1 + 49] = 25 ¥ 50 = 625. 2 2 2 Hence, the required sum of odd numbers between 0 and 50 is 625. EXAMPLE 16. Find the sum of all three digit numbers which leave the remainder 3 when divided by 5. SOLUTION. Here, we have The least 3-digit number which leaves the remainder 3 on division by 5 = 103 The largest 3-digit number which leaves the remainder 3 on division by 5 = 998 Thus, the list of such 3-digit numbers is 103, 108, 113, ..............., 993, 998. ...(1) Now, T2 - T1 = 108 - 103 = 5 and T3 - T2 = 113 - 108 = 5  T2 - T1 = T3 - T2 So, the list of numbers in (1) forms an A.P with a = 103, d = 5 and l = 998 Using the General term formula, we have Tn = a + (n - 1)d fi 998 = 103 + (n - 1)5 fi 5(n - 1) = 895 fi 5(n - 1) = 998 - 103 fi n-1= 895 = 179 5 fi n = 179 + 1 = 180 Thus, we need to find the sum of 180 3-digit numbers which leave the remainder 3 when divided by 5. Using the Sum formula of an A.P, we get Sn = n [a + l] 2 fi S180 = 180 [103 + 998] = 90 [1101] = 99090 2 Hence, the required sum of all three digit number each of which leaves remainder 3 when divided by 5 is 99090. EXAMPLE 17. Find the sum of all two digit natural numbers greater than 50 which when divided by 7 yield a remainder of 4. SOLUTION. The two digit natural numbers greater than 50 which when divided by 7 leave a remainder 4 are: 53, 60, 67, ..............., 95 ...(1) Now, T2 - T1 = 60 - 53 = 7 and T3 - T2 = 67 - 60 = 7  T2 - T1 = T3 - T2 So, the above list of numbers forms an A.P, with a = 53, d = 7 and l = 95 Let the total number of terms of the A.P: (1) be n. Using the General term formula of an A.P, we have

64 MATHEMATICS-X Tn = a + (n – 1) d fi l = a + (n – 1) d fi 95 = 53 + (n – 1) 7 fi (n – 1)7 = 95 – 53 = 42 fi (n – 1) = 42 =6 7 fi n=6+1=7 We need to find S7 of the above A.P. Using the Sum Formula of an A.P, we have Sn = n [a + l] 2 fi S7 = 7 (53 + 95) = 7 ¥ 148 = 7 ¥ 74 = 518 2 2 Hence, the required sum of all two digit natural numbers greater than 50 which when divided by 7 yield a remainder of 4 is 518. EXAMPLE 18. Subba Rao started work in 1995 at an annual salary of ` 5000 and received an increment of ` 200 each year. In which year did his income reach ` 7000? [NCERT, CBSE 2012] SOLUTION. Here, we have Subba Rao’s annual salary at the start, a = ` 5000 Subba Rao’s increment each year, d = ` 200 Let n be the year in which his income reach ` 7000. Then, Subba Rao’s salary in the last (nth) year, Tn = ` 7000 Using the General term formula, we have Tn = a + (n - 1)d fi Tn = 5000 + (n - 1) 200 fi (n - 1) = 2000 200 fi 7000 = 5000 + (n - 1) 200 fi n - 1 = 10 fi (n - 1) 200 = 7000 - 5000 fi n = 10 + 1 fi (n - 1) 200 = 2000 fi n = 11 So, Subba Rao’s income would reach ` 7000 in the 11th year (i.e., 2005). EXAMPLE 19. Jasleen saved ` 5 in the first week of the year and then increased her weekly savings by ` 1.75. In what week, will her weekly savings be ` 20.75? [NCERT] SOLUTION. Here, we have a =`5 Jasleen’s Savings in the first week, d = ` 1.75 Jasleen’s increase in weekly savings,

ARITHMETIC PROGRESSIONS 65 Let in the nth week her weekly savings become ` 20.75 Then, Jasleen’s saving in the nth (last) week, Tn = ` 20.75 Using the General term formula, we have Tn = a + (n - 1)d fi Tn = 5 + (n - 1) 1.75 fi 20.75 = 5 + (n - 1) 1.75 fi (n - 1) 1.75 = 20.75 - 5 = 15.75 fi n - 1 = 15.75 = 9 1.75 fi n = 9 + 1 = 10 Hence, Jasleen’s weekly savings will be ` 20.75 in 10th week. TEST YOUR KNOWLEDGE 1. The sum of the 5th and 9th terms of an A.P is 30. If its 25th term is three times its 8th term, find the A.P. [CBSE 2014] 2. Find the value of n for which the nth terms of two A.P.s: 9, 7, 5, ............... and 15, 12, 9, ............... are equal. [CBSE 2015] 3. The eighth term of an A.P is half its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term. [NCERT (EP)] 4. Find the middle term of the A.P. 7, 13, 19, ..............., 247. [CBSE 2020] 5. If the mth term of an A.P is 1 and nth term is 1 , then show that its (mn)th term is 1. nm [CBSE 2019] 6. The first term of an A.P. of 20 terms is 2 and its last term is 59. Find its 6th term from the end. [CBSE 2016] 7. In an A.P, the first term is 25, nth term is –17, and sum of first n terms is 60. Find n and d, the common difference. [CBSE 2012] 8. The sum of first six terms of an arithmetic progression is 42. The ratio of its 10th term to its 30th term is 1 : 3. Calculate the first term and the thirteenth term of the A.P. [CBSE 2012] 9. The sum of the first seven terms of an A.P is 182. If its 4th and the 17th terms are in the ratio 1 : 5, find the A.P. [CBSE 2014] 10. If the sum of first m terms of an A.P. is the same as the sum of its first n terms, show that the sum of its first (m + n) terms is zero. [CBSE 2019 ] 11. Find the common difference of an A.P whose first term is 5 and the sum of its first four terms is half the sum of the next four terms. [CBSE 2012] 12. The sum of n, 2n, 3n terms of an A.P are S1, S2, S3 respectively. Prove that S3 = 3 (S2 – S1). [CBSE 2012] 13. The sum of the first 8 terms of an A.P is 100 and the sum of its first 19 terms is 551. Find the first term and the common difference. [CBSE 2012] 14. The first term of an A.P is 25; sum of the first n terms being 60 and nth term is –17. Find the tenth term. [CBSE 2016]

66 MATHEMATICS-X 15. If the mth term of an A.P is 1 and the nth term is 1 , show that the sum of (mn) terms n m is 1 (mn + 1). [CBSE 2015, 17] 2 16. Find the sum of all eleven terms of an A.P whose middle term is 30. [CBSE 2020] 17. Find the sum of all odd numbers up to 85. [CBSE 2016] 18. Find the sum of all two digit natural numbers which when divided by 3 yield 1 as remainder. [CBSE 2012] 19. (i) Find the sum of all multiples of 8 lying between 201 and 950. [CBSE 2012] (ii) Find the sum of all multiples of 7 lying between 500 and 900. [CBSE 2012] (iii) Find the sum of all multiples of 9 lying between 400 and 800. [CBSE 2012] 20. (i) How many two-digit numbers are divisible by 3? [CBSE 2012] (ii) How many three-digit numbers are (a) divisible by 11? (b) divisible by 12? [CBSE 2012] (iii) Find the number of all three-digit numbers which are divisible by 9. [CBSE 2013] (iv) Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5. [CBSE 2014] (v) Find the number of natural numbers between 102 and 998 which are divisible by 2 and 5 both. [CBSE 2019] 21. The sum of the 5th and the 9th terms of an A.P is 30. If its 25th term is three times its 8th term, find the A.P. [CBSE 2014] 22. Find the sum of all 3-digit natural numbers which are multiples of 11. [CBSE 2012] 23. Solve the equation: 1 + 4 + 7 + 10 + ... + x = 287 [CBSE 2020] 24. Show that the sum of all terms of an A.P. Whose first term is a, the 2nd term in b and the last term is c is equal to (a + c)(b + c – 2a) . 2(b – a) 25. Find a, b and c, if it is given that the number a, 7, b, 23, c are in A.P. [CBSE 2020] 26. For an A.P., it is given that the first term (a) = 5, common difference (d) = 3, and the nth term (an) = 50. Find n and the sum of first n terms of A.P. [CBSE 2020] LONG ANSWER TYPE QUESTIONS Illustrative Examples EXAMPLE 1. Find five numbers in an A.P whose sum is 12 1 and ratio of the first to the last term 2 is 2 : 3. [CBSE 2014] SOLUTION. Let the required numbers be a – 2d, a – d, a, a + d, a + 2d ...(1) Sum of these five numbers, S = (a - 2d) + (a - d) + a + (a + d) + (a + 2d) = 5a fi 12 1 = 5a fi 5a = 25 2 2

ARITHMETIC PROGRESSIONS 67 fi a= 25 ¥ 1 = 5 ...(2) 2 5 2 According to the problem, a – 2d 2 5 – 2d 2 a + 2d 3 2 3 = fi 5 = Using (2) 2 2d Cross multiplying + ...(3) fi 3 È 5 – 2d˘˙˚ = 2 È 5 + 2d˚˙˘ ÍÎ 2 ÎÍ 2 fi 15 – 6d = 5 + 4d fi 10d = 15 – 5= 5 2 2 2 fi d = 5 ¥ 1 = 1 2 10 4 Putting the values from (2) and (3) in (1), we get È5 – 2 Ê 1 ˆ ˘ , È5 – 1 ˘ , 5 , È5 + 1 ˘ , È5 + 2 Ê 1 ˆ ˘ i.e., 2, 9 , 5 , 11 , 3 ÎÍ 2 ËÁ 4 ¯˜ ˚˙ ÎÍ 2 4 ˚˙ 2 ÎÍ 2 4 ˚˙ ÎÍ 2 ËÁ 4 ¯˜ ˚˙ 4 2 4 Hence, the required five numbers in A.P are: 2, 9 , 5 , 11 and 3. 4 2 4 EXAMPLE 2. The sum of first six terms of an A.P is 42. The ratio of its 10th term to 30th term is 1 : 3. Calculate the first and 13th term of the A.P. SOLUTION. Let the first term and the common difference of the given A.P be ‘a’ and ‘d’ respectively. Then, T10 = a + (10 – 1)d = a + 9d ...(1) and T30 = a + (30 – 1)d = a + 29d ...(2) \\ T10 = a + 9d Using (1) and (2) T30 a + 29d fi 1 = a + 9d   T10 : T30 = 1 : 3 (Given) 3 a + 29d fi 3a + 27d = a + 29d. Cross-multiplying fi 2a = 2d fi a=d ...(3) Now, S6 = 6 [2a + 5d]  Sn = n {2a + (n – 1)d} fi 2 2 = (6a + 15d) = (6a + 15a)   a = d by (3) 42 = 21a fi 21a = 42 fi a= 42 =2 21 Hence, the first term is 2 and the 13th term is 26.

68 MATHEMATICS-X EXAMPLE 3. The sum of four consecutive numbers in an A.P is 32 and the ratio of the product of the first and the last terms to the product of the two middle terms is 7 : 15. Find the numbers. [NCERT (EP), CBSE 2018] SOLUTION. Let the required four consecutive numbers in A.P be (a - 3d), (a - d), (a + d) and (a + 3d). Sum of these numbers, S = (a - 3d) + (a - d) + (a + d) + (a + 3d) = 4a ... (1) Also, S = 32 Given fi 4a = 32 Using (1) fi a=8 ... (2) According to the problem, Product of (I) and (IV) = 7 Product of (II) and (III) 15 fi (a - 3d)(a + 3d) = 7 fi a2 - 9d2 = 7 (a - d)(a + d) 15 a2 - d2 15 fi 64 - 9d2 = 7 Using (2) 64 - d2 15 fi 960 – 135d2 = 448 – 7d2 Cross multiplying fi – 128d2 = 448 – 960 = –512 fi 128d2 = 512 fi d2 = 512 =4 fi d=±2 128 Hence, a = 8 and d = 2. Hence, the four terms are : (a - 3d), (a - d), (a + d) and (a + 3d) Case (I): When d = 2. The numbers are [8 - 3(2)], (8 - 2), (8 + 2) and [8 + 3(2)] i.e., 2, 6, 10, 14. Case (II): When d = –2. The numbers are [8 - 3(–2)], (8 - (–2), (8 – 2) and [8 + 3(–2)] i.e., 14, 10, 6, 2. Hence, the required four numbers are 2, 6, 10, 14 or 14, 10, 6, 2. EXAMPLE 4. If the sum of first 7 terms of an A.P is 49 and that of 17 terms is 289, find the sum of first n terms. SOLUTION. Let a be the first term and d be the common difference of an A.P. Sum of the first 7 terms of the A.P, S7 = 49 Sum of the first 17 terms of the A.P, S17 = 289 Sum of the first n terms of the A.P, Sn = ? Using the Sum formula of an A.P, we get Sn = n [2(a) + (n - 1)d] i.e., S7 = 7 [2a + (7 - 1)d] 2 2

ARITHMETIC PROGRESSIONS 69 fi 49 = 7 [a + 3d] fi a + 3d = 7 ...(1) and S17 = 17 [2a + (17 - 1)d] fi 289 = 17 [a + 8d] 2 fi a + 8d = 289 = 17 ...(2) 17 Subtracting (1) from (2), we get 8d - 3d = 17 - 7 fi 5d = 10 fi d= 10 =2 5 Putting d = 2 in (1), we get a + 3(2) = 7 fi a = 7 - 6 = 1 \\ Sn = n [2(a) + (n - 1)d] 2 = n [2(1) + (n - 1)2] Putting a = 1 and d = 2 2 = n[1 + n - 1] = n2 Hence, the required sum of first n terms of the given A.P is n2. EXAMPLE 5. If the pth term of an A.P is 1 , and the qth term is 1 , show that the sum of pq q p terms is ( pq + 1) . [CBSE 2012] 2 SOLUTION. Let a be the 1st term and d be the common difference of the A.P then, according to the problem: Tp = 1 fi a + (p – 1)d = 1 ...(1) q q and Tq = 1 fi a + (q – 1)d = 1 ...(2) p p Subtracting (2) from (1), we have d(p – q) = 1 - 1 = p-q fi d= p-q ¥ p 1 q = 1 ...(3) q p pq pq - pq Putting the value of d from (3) in (1), we have a + (p – 1) 1 = 1 fi a = 1 – p-1 pq q q pq fi a= p- p +1 = 1 ...(4) pq pq \\ Sum of pq terms = pq ÈÍ2 Ê 1 ˆ + (pq - 1) 1 ˘  Sn = n ÎÈ2a + (n - 1)d˘˚ 2 ÍÎ ËÁ pq ˜¯ pq ˙ 2 ˚˙

70 MATHEMATICS-X = pq ◊ 1 [2 + pq - 1] = (pq + 1) 2 pq 2 Hence, proved. EXAMPLE 6. The sum of first 20 terms of an A.P is 400 and the sum of first 40 terms is 1600. Find the sum of its first 10 terms. [CBSE 2015] SOLUTION. Here, we have S20 = 400 and S40 = 1600 Let the 1st term and the common difference of the given A.P. be ‘a’ and ‘d’ respectively Then, S20 = 20 [2a + (20 – 1)d] Using the Sum formula fi 2 400 = 10(2a + 19d) fi 2a + 19d = 40 ...(1) and S40 = 40 [2a + (40 – 1)d] fi 1600 = 20[2a + 39d] 2 fi 2a + 39d = 80 ...(2) Subtracting (1) from (2), we get 39d – 19d = 80 – 40 fi 20d = 40 fi d = 40 = 2 ...(3) 20 Putting the value of ‘d’ from (3) in (1), we get 2a + 19(2) = 40 fi 2a = 40 – 38 = 2 fi a = 1. ...(4) \\ S10 = 10 [2(1) + (10 – 1)(2)]  Sn = n [2a + (n - 1)d] 2 2 = 5[2 + 18] = 5 ¥ 20 = 100 Hence, the required sum of first ten terms of the given A.P is 100. EXAMPLE 7. Solve the equation: 1 + 4 + 7 + 10 + ............... + x = 287. SOLUTION. Here, the terms in LHS of the given equation, namely, 1, 4, 7, 10, ..............., x form an A.P with a = 1, d = 3, l = x Then, x = 1 + (n – 1)3   l = a + (n – 1)d fi x = 3n – 2 ...(1) Also, S = n (a + l) fi 287 = n (1 + x) 2 2 fi 287 = n (1 + 3n – 2) Using (1) 2 3n2 – n – 574 = 0 fi 574 = n (3n – 1) fi which is a quadratic equation in n.

ARITHMETIC PROGRESSIONS 71 fi n = 1± 1 + 6888 = 1 ± 83 = 84 , - 82 = 14, - 41 6 6 6 6 3 = 14  n π - 41 (being – ve) 3 Putting the value of n = 14 in (1), we have x = 3n – 2 = 3(14) – 2 = 42 – 2 = 40. EXAMPLE 8. If in an A.P., the sum of first m terms is n and the sum of its first n terms is m, the prove that the sum of its first (m + n) terms is – (m + n) [CBSE 2020] SOLUTION. Let a be the 1st term and d be see common difference of the given A.P. Then, according to the problem: First condtion Second condition Sm = n Sn = m fi m ÎÈ2a + (m - 1) d˘˚ =n fi n ÎÈ2a + (n - 1) d˘˚ =m 2 2 fi 2am + m(m – 1)d = 2n ...(1) 2an + n(n – 1)d = 2m ...(2) ...(3) Subtracting (2) from (1), we get 2a(m – n) + {m (m – 1) – n(n – 1)}d = 2n – 2m fi 2a(m – n) + {m2 – m – n2 + n}d = 2n – 2m fi 2a(m – n) + {(m2 – n2) – (m – n)}d = –2 (m – n) fi 2a(m – n) + {(m – n) {(m + n) – 1}d = –2 (m – n) Dividing throughout by (m – n), we get 2a + (m + n – 1)d = –2 Now, Sm+n = m + n ÎÈ2a + (m + n - 1) d˚˘ 2 fi Sm+n = m + n ¥ (–2) Using (3) 2 fi Hence, proved Sm+n = –(m + n) EXAMPLE 9. The sum of first q terms of an A.P is 63 q – 3q2. If its pth term is – 60, find the value of p. Also, find the 11th term of this A.P. [CBSE 2013] SOLUTION. Here, we have ...(1) Tp = –60 and Sq = 63q – 3q2 Changing q to (q – 1) in (1), we get Sq – 1 = 63(q – 1) – 3(q – 1)2 = 63q – 63 – 3(q2 – 2q + 1) ...(2) = –3q2 + 69q – 66 Tq = Sq – Sq – 1 Using (1) and (2) = (63q – 3q2) – (–3q2 + 69q – 66) = 63q – 3q2 + 3q2 – 69q + 66 = 66 – 6q

72 MATHEMATICS-X fi Tp = 66 – 6p ...(3) Putting Tp = – 60 (Given) in (3), we get – 60 = 66 – 6p fi 6p = 66 + 60 = 126 fi p= 126 = 21 To find T11, we put p = 11 in (3) 6 \\ T11 = 66 – 6 ¥ 11 = 66 – 66 = 0. EXAMPLE 10. A sum of ` 700 is to be used for giving 7 cash prizes to students of a school for their academics performance. If each prize is ` 20 less than its preceding prize, find the value of each of the prizes. [CBSE 2012] SOLUTION. Here, we have Total value of 7 cash prizes, S7 = ` 700 Difference of any 2 consecutive prizes, d = ` 20 Suppose, the first prize, = ` a Number of Prizes, n = 7 A.P = ? Using the Sum formula of an A.P, we have Sn = n [2a + (n - 1)d] 2 fi S7 = 7 [2a + (7 - 1)20] 2 fi 700 = 7 [2a + 6¥ 20] = 7 È 2a + 6 ¥ 20 ˘ 2 ÍÎ 2 2 ˙˚ = 7[a + 60 ] = 7a + 420 fi 7a = 700 - 420 = 280 fi a= 280 = 40 7 \\ A.P is a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, a + 6d i.e., 40, 40 + 20, 40 + 2(20), 40 + 3(20), 40 + 4(20), 40 + 5(20), 40 + 6(20) 40, 60, 80, 100, 120, 140, 160. \\ Value of the prizes are 160, 140, 120, 100, 80, 60, 40. TEST YOUR KNOWLEDGE 1. The sum of first m terms of an A.P is 4m2 – m. If its nth term is 107, find the value of n. Also, find the 21st term of this A.P. [CBSE 2013] 2. The sum of first n terms of an A.P is 5n2 + 3n. If its mth term is 168, find the value of m. Also, find the 20th term of the A.P. [CBSE 2013] 3. If Sn denotes the sum of first n terms of an A.P., prove that S12 = 3 (S8 – S4). [NCERT (EP) CBSE 2021 (C)]

ARITHMETIC PROGRESSIONS 73 4. Ram saved ` 15 in the first week of a year and then increased his weekly savings by ` 2.50. If in the nth week, his savings become ` 22.50, find n. [CBSE 2015] 5. Find the sum of all natural numbers between 200 and 700 which are odd and divisible by 7. [CBSE 2016] 6. Find an A.P, whose fourth term is 9 and the sum of its sixth and thirteenth term is 40. 7. The sum of the first 30 terms of an A.P. is 1920. If the 4th term is 18, Find its 11th term. [CBSE 2020] 8. If Sn denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 – S4) [CBSE 2021] CASE STUDY/SOURCE BASED INTEGRATED MCQS CASE STUDY 1 India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year. Based on the above information, answer the following questions: 1. Find the production during first year. 2. Find the production during 8th year. 3. Find the production during first 3 years. 4. In which year, the production is ` 29,200. 5. Find the difference of the production during 7th year and 4th year. CASE STUDY 2 Your friend Veer wants to participate in a 200 m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to do in 31 seconds.

74 MATHEMATICS-X 1. Which of the following terms are in A.P. for the given situation (a) 51, 53, 55….... (b) 51, 49, 47….... (c) –51, –53, –55….... (d) 51, 55, 59….... 2. What is the minimum number of days he needs to practice till his goal is achieved (a) 10 (b) 12 (c) 11 (d) 9 3. Which of the following term is not in the A.P. of the above given situation (a) 41 (b) 30 (c) 37 (d) 39 4. If nth term of an A.P. is given by an = 2n + 3 then common difference of an A.P. is (a) 2 (b) 3 (c) 5 (d) 1 5. The value of x, for which 2x, x + 10, 3x + 2 are three consecutive terms of an A.P. (a) 6 (b) –6 (c) 18 (d) –18 CASE STUDY 3 Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of ` 1,18,000 by paying every month starting with the first instalment of ` 1000. If he increases the instalment by ` 100 every month, answer the following: 1. The amount paid by him in 30th installment is (a) 3900 (b) 3500 (c) 3700 (d) 3600 2. The amount paid by him in the 30 installments is (a) 37000 (b) 73500 (c) 75300 (d) 75000 3. What amount does he still have to pay offer 30th installment? (a) 45500 (b) 49000 (c) 44500 (d) 54000 4. If total installments are 40 then amount paid in the last installment? (a) 4900 (b) 3900 (c) 5900 (d) 9400 5. The ratio of the 1st installment to the last installment is (a) 1:49 (b) 10:49 (c) 10:39 (d) 39:10

ARITHMETIC PROGRESSIONS 75 Answers VERY SHORT ANSWER TYPE QUESTIONS 1. k = 3 (ii) 35th term (iii) Not 2. (i) 8th term (ii) -16 (iii) 0 (ii) 90 3. (i) 209 4. (i) -979 5. (i) Sn = n(n + 1) (ii) 400 (iii) 500500 2 6. (i) 30 (ii) No, as common difference does not exist 7. n = 10 8. a = 28 23 10. a = 45 9. a = 10 SHORT ANSWER TYPE-I QUESTIONS 1. 17 2. 13th term 3. (i) 81 (ii) 75 4. (i) 3, 4, 5, 6, 7, ............... (ii) 5, 1, -3, -7 (iii) 8, 12, 16, 20, ............... 5. (i) 25th term (ii) 49th term 6. (i) 53rd term (ii) 31st term 7. (i) please give answer 8. (i) 3 (ii) 3 11. 4 or 13 12. 1188 10. 108 16. x = 101 9. 672 15. –44 13. 32nd term SHORT ANSWER TYPE-II QUESTIONS 1. 3, 5, 7, 9, ............... 2. n = 7 3. 3 4. 127 8. 1st term = 2, 13th term = 26 6. 44. 7. n = 15, d = –3. 9. 2, 10, 18, 26, ............... 11. d = 2. 13. a = 2, d = 3 14. –2. 16. 330 17. 1849 18. 1605 19. (i) 53568 (ii)39900 (iii) 26334 20. (i) 30 (ii) (a) 81 (b) 75 (iii) 100 (iv) 89 (v) 89 21. 3, 5, 7, ............... 22. 44550 23. x = 40 25. a = –1, b = 15 and c = 31 26. n = 16, Sn = 440 LONG ANSWER TYPE QUESTIONS 1. n = 14; T21 = 163 2. m = 17; T20 = 198 4. 4 5. 16128 6. 3, 5, 7, 9, ............... 7. a11 = 46

76 MATHEMATICS-X CASE STUDY 1. 1. ` 5000 2. Production during 8th year is (a + 7d) = 5000 + 2(2200) = 20400 3. Production during first 3 year = 5000 + 7200 + 9400 = 21600 4. N = 12 5. Difference = 18200 – 11600 = 6600 2. 1. (b) 2. (c) 3. (b) 4. (a) 5. (a) 3. 1. (a) 3900 2. (b) 73500 3. (c) 44500 4. (a) 4900 5. (b) 10 : 49

6CHAPTER SURFACE AREAS AND VOLUMES IMPORTANT POINTS TO REMEMBER FOR QUICK REVISIONS In this chapter, we will study the following points to remember. ∑ Combination of Solids: When any two of the basic solids, namely, cuboid, cone, cylinder and sphere are combined, a new solid is formed. ∑ Surface Areas of Combination of Solids: The total surface areas of combinations of solids are obtained by adding together the exposed portions of the surface areas of the individual solids, and not by adding together their total surface areas. ∑ Volumes of Combinations of Solids: The volumes of combinations of solids are actually the sum of the volumes of the constituents forming the new solids. ∑ Conversion of Solids: When one solid is converted into another, the volume of material in the original solid equals the volume of material in the second. ∑ Frustum of a Right Circular Cone: When a right circular cone is cut (or sliced through) by a plane parallel to its base, through some points on its axis and the smaller conical portion containing the vertex is removed, then the resulting left out portion of the solid, (which is in the form of a bucket) is called a Frustum of a Right Circular Cone. It consists of two unequal flat circular bases and a curved surface. ∑ Formulae for the Frustum of a Cone: The formulae involving the frustum of a cone are: ( )V = 1 (i) Volume of the Frustum of a Cone, 3 ph R2 + r2 + Rr (ii) Curved Surface Area of the Frustum of a Cone, S = p l (R + r) where l = h2 + (R - r)2 St = [p R2 + pr2 + p (R + r)] = pl (R + r) + pR2+ pr2 (iii) Total Surface Area of the frustum of a cone, where l = h2 + (R - r)2 152

SURFACE AREAS AND VOLUMES 153 (iv) Total Surface Area of the frustum of a cone which is open at one of the bases, St = pl (R + r) + pr2 or St = pl (R + r) + p(R2 + 7pr2), Depending on the end which is open 1. The capacity of a bucket is equal to its volume. 2. The above formulae for the frustum of a cone can also be used for a bucket with R > r. ∑ Standard Form of a Quadratic Equation: The most general form of a quadratic equation in the variable x called the standard form, is: ax2 + bx + c = 0, where a π 0, a, b, c are real numbers. ∑ Roots (or Solutions) of a Quadratic Equation: Those values of x, which satisfy a quadratic equation, are called roots (or solutions) of the equation. Thus, a real number a is called a root of the quadratic equation ax2 + bx + c = 0, if aa2 + ba + c = 0. A quadratic equation can have at most two roots, which are usually denoted by a and b. The zeores of the quadratic polynomial ax2 + bx + c, and the roots of the quadratic equation ax2 + bx + c = 0, are the same. ∑ Discriminant of a Quadratic Equation: It is a relationship between the coefficients of a quadratic equation and is given by Disc. D = b2 – 4ac. ∑ Finding the Roots of a Quadratic Equation: There are three methods to find the roots of a quadratic equation ax2 + bx + c = 0, a π 0. (i) By the Factorisation Method: It is applied when the discriminant of a quadratic equation, D(= b2 – 4ac) is a perfect square of a positive number. In this case, factorise ax2 + bx + c, a π 0, into a product of two linear factors, and then the roots of the quadratic equation ax2 + bx + c = 0 are found by equating each factor to zero. (ii) By the Method of Completing the Squares: The key point involved in its 5-Step Procedure is the addition of [ 1 the coefficient of x]2 on both the sides, with 2 leading coefficient unity and constant term to R.H.S. (iii) By the Quadratic Formula: It directly gives the two roots of a Quadratic Equation, provided its discriminant D ≥ 0, by the formula stated below: x= -b ± D, where D = b2 – 4ac 2a ∑ Nature of Roots of a Quadratic Equation: The nature of roots of a quadratic equation ax2 + bx + c = 0, depends upon the nature of its discriminant. (i) If D > 0, (i.e., positive), then the roots are real and unequal/distinct. (ii) If D = 0, then the roots; are real and equal/coincident. (iii) If D < 0, then there are no real roots.

154 MATHEMATICS-X PART A VERY SHORT ANSWER TYPE QUESTIONS Illustrative Examples EXAMPLE 1. Three equal cubes are placed side by side in a row. Find the ratio of the surface area of the new cuboid so formed to the sum of the surface areas of the 3 cubes. SOLUTION. Let the edge of each cube be ‘a’ Then, surface area of each cube = 6a2 fi Sum of the surface areas of 3 cubes ...(1) S1 = 3 ¥ (6a2) = 18a2 Dimensions of the new cuboid = l ¥ b ¥ h = 3a ¥ a ¥ a ...(2) \\ Surface area of the new cuboid S2 = 2(lb + bh + hl) = 2[3a ◊ a + a ◊ a + a ◊ 3a] = 2(7a2) = 14a2 From (1) and (2), we have S2 : S1 = S2 14 a2 = 7 = 7: 9 S1 = 18 a2 9 Hence, the required ratio of the surface area of the new cuboid to the sum of the surface areas of 3 cubes is 7 : 9. EXAMPLE 2. A solid cone of radius r and height h is placed over a solid cylinder having same base-radius and height as that of a cone. Find the total surface of the combined solid. SOLUTION. TSA of the combined solid = (CSA of the cone) + (CSA of the solid cylinder) = prl + 2prh + pr2 = pr (l + 2h + r) = pr È r2 + h2 + 2h + r˚˙˘  l = r2 + h2 ÍÎ Hence, the required TSA of the combined solid is pr ( )r2 + h2 + 2h + r units2. EXAMPLE 3. Two identical solid hemisphere of equal base-radius, r cm, are struck together along their bases. Find the total surface area of the combination. SOLUTION.  CSA of a hemisphere = 2pr2 Fig. 6.1 fi TSA of the combination = 2pr2 + 2pr2   Base of both the hemispheres is common = 4pr2 and that has been joined so ignore it. Hence, the required TSA of the combination is 4pr2.

SURFACE AREAS AND VOLUMES 155 EXAMPLE 4. Three cubes, whose edges measure 3 cm, 4 cm, and 5 cm are respectively form a single cube. Find volume of the new cube. SOLUTION. Volume of the new cube, V = x3 cm3 = (6 ¥ 6 ¥ 6) cm3 = 216 cm3. Hence, the required volume of the new cube is 216 cm3. EXAMPLE 5. Three cubes each of side 5 cm are joined end-to-end. Find the volume of the resulting cuboid. SOLUTION. When three cubes are joined end-to-end, the cuboid is formed (as shown in Fig. 6.2). For the resulting cuboid, we have Length, l = (5 + 5 + 5) cm = 15 cm Breadth, b = 5 cm Height, h = 5 cm Volume of the resulting cuboid, Fig. 6.2 V = (l ¥ b ¥ h) cm3 = (15 ¥ 5 ¥ 5) cm3 = 375 cm3. Hence, the required volume of the resulting cuboid is 375 cm3. EXAMPLE 6. Three cubes whose edges measure 3 cm, 4 cm and 5 cm respectively to form a single cube. Find its edge. SOLUTION. Let x cm be the edge of the new cube. Then, Volume of the new cube = Sum of the volume of three cubes. x3 = 33 + 43 + 53 = 27 + 64 + 125 fi x3 = 216 fi x3 = 63 fi x = 6 cm   a3 = b3 fi a = b Thus, edge of the new cube is 6 cm long. EXAMPLE 7. Find the maximum volume of a cone that can be curved out of a solid hemisphere of radius r. A rO SOLUTION. Here, base-radius of the cone B = Radius of the hemisphere = r r Height of the cone = Radius of the hemisphere \\ Volume of the cone = 1 pr 2 ¥ r = 1 pr 3 units3. V 3 3 Fig. 6.3

156 MATHEMATICS-X EXAMPLE 8. Find the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube. SOLUTION. Let the radius of the sphere which fits exactly into a cube be r units. Then length of edge of the cube = 2r units Let V1 and V2 be the volumes of the cube and sphere respectively. Then, V1 = (2r)3 = 8r3 and V2 = 4 pr 3 3 \\ V1 = 8r3 = 6 =6: p V2 p 4 pr3 3 Hence, the required ratio of the volume of a cube to that of a sphere is 6 : p. EXAMPLE 9. Find the capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Fig. 6.4. SOLUTION. We know that: Capacity of a cylindrical vessel, C1 = pr2h units3 and Capacity of a hemisphere, C2 = 2 pr 3 units3 3 \\ Capacity of the vessel, C = C1 – C2 = Ê pr 2 h - 2 pr 3 ˆ units3 Fig. 6.4 ÁË 3 ˜¯ = 1 pr 2 (3h – 2r) units3 3 EXAMPLE 10. Volumes of two spheres are in the ratio 27 : 64. What is the ratio of their surface areas? [CBSE 2015] SOLUTION. Let r1 and r2 be the radii of the two spheres respectively. Then V1 = 4 pr13 = Ê r1 ˆ 3 V2 3 ÁË r2 ¯˜ 4 3 pr23 fi 27 = Ê r1 ˆ 3 fi Ê r1 ˆ 3 = Ê 3ˆ3 64 ËÁ r2 ˜¯ ÁË r2 ¯˜ ÁË 4˜¯ fi r1 = 3 ...(1) r2 4 Using (1) \\ S1 = 4pr12 = Ê r1 ˆ 2 = Ê 3ˆ2 = 9 S2 4pr22 ÁË r2 ˜¯ ÁË 4 ˜¯ 16 \\ S1 : S2 = 9 : 16 Hence, the required ratio of the surfaces is 9 : 16.

SURFACE AREAS AND VOLUMES 157 EXAMPLE 11. A cylinder and a cone are of same base-radius and of same height. Find the ratio of the volume of cylinder to that of the cone. [CBSE 2015] SOLUTION. Let the common base-radius be r cm and the common height be h cm. Then, volume of the cylinder, V1 = pr2h units3 and the volume of the cone, V2 = 1 pr2h units3 3 \\ V1 : V2 = V1 = pr 2 h = 3 = 3:1 V2 1 1 pr 2 h 3 Hence, the required ratio of the volumes is 3 : 1. EXAMPLE 12. The radii of the bases of two solid right circular cones of same height cones are r1 and r2 respectively. The cones are melted and recast into a cylinder. Find the base-radius of the cylinder. SOLUTION. Let the radius of the cylinder be R units, and the height of the 2 cones be h units. Then Volume of the cylinder = Sum of the volumes of two cones fi pR2h = 1 pr21h + 1 pr22h fi phR2 = 1 ph(r21 + r22) 3 3 3 fi R2 = 1 (r21 + r22) Cancelling ph on both sides 3 fi R= r12 + r22 is the required radius 3 EXAMPLE 13. A solid metallic hemisphere of radius 4 cm is melted and recasted into a solid right circular cone of base diameter 8 cm. What is the height of this cone? [CBSE 2015] SOLUTION. Let the height of the cone be h cm. 8 cm Base-diameter of the cone = 8 cm fi Base-radius of the cone, r= 8 = 4 cm 2 Radius of the hemisphere, r¢ = 4 cm Then, volume of the hemisphere = Volume of the solid cone Fig. 6.5(i) fi 2 pr ¢ 3 = 1 pr2h 4 cm 3 3 Fig. 6.5(ii) fi 2 ¥ (4)3 = 1 ¥ (4)2 ¥ h 3 3 fi h= 2 ¥ 3 ¥ (4)3 = 8 cm 3 ¥ (4)2 Hence, the required height of the cone is 8 cm.

158 MATHEMATICS-X TEST YOUR KNOWLEDGE 1. (i) Two cubes each of 10 cm edge are joined end-to-end. Find the surface area of the resulting cuboid. (ii) Three cubes, each of side 5 cm, are joined end-to-end to form a cuboid. Find the surface area of the cuboid. (iii) A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. Find the total surface area of the shape so formed. (iv) The radii of two cylinders are in the ratio 3 : 5 and their heights are in the ratio 2 : 3. What is the ratio of their curved surface areas? (v) The surface area of the sphere is same as the CSA of a right circular cylinder whose height and diameter are 12 cm each. Find the radius of the sphere. 2. (i) Three cubes of side 4 cm are joined to form a cuboid. Find the volume of the resulting cuboid. (ii) Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 18 m. (iii) Find the volume of the largest sphere that can be carved out of a cube of a side 7 cm. (iv) A solid ball is exactly fitted inside the cubical box of side ‘a’. Find the volume of the ball. (v) Find the volume of the largest right circular cone that can be cut off from a cube of edge 4.2 cm. 3. (i) Three cubes of metal, whose edges are 6 cm, 8 cm and 10 cm, are melted and one new cube is made. Find the total surface area of the new cube. (ii) The dimensions of a metallic cuboid are 100 cm ¥ 80 cm ¥ 64 cm. It is melted and recast into a cube. Find the surface area of the cube. (iii) By melting a solid sphere of radius 5 cm, a solid right circular cone of the same circular base is made. Find the height of the cone. (iv) The radii of the bases of two circular solid cones of same height are r1 and r2 respectively. The cones are melted and recast into a solid sphere of radius R. Find the height of each cone. (v) How many steel rods, each of length 7 m and diameter 2 cm can be made of 0.88 cubic metre of steel? 4. 12 solid spheres of the same radii are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. Find the diameter of each sphere. [CBSE (SP)’ 20] 5. Two cones have their heights in the ratio 1 : 3 and radii in the ratio 3 : 1. What is the ratio of their volumes? [CBSE 2020] PART B SHORT ANSWER TYPE-I QUESTIONS Illustrative Examples EXAMPLE 1. Two cubes each of volume 64 cm3 are joined end-to-end. Find the surface area of the resulting cuboid. [NCERT, CBSE 2012] SOLUTION. Since, the volume of the cube = (Edge)3