The Leading Maths - 9

Algebra Algebra Apply the algebraic knowledge, skill and relationship on the solution of the behaviour problems. Estimated Working Hours : Competency Learning Outcomes At the end of this unit, the students will be able to: find the general terms of sequence and series according to their patterns and express them in sigma notation ( ∑ ). solve the problems related to arithmetic and geometric sequences. factorise the algebraic expressions if the form of (a ± b)3 , a3 ± b3 and a4 + a2 b2 + b4 . find the highest common factor (HCF) and lowest common multiple (LCM) of the algebraic expressions. solve the problems on daily life activities related to simultaneous equations of two variables. develop the skill of using the laws of indices in simplification of algebraic expressions. 8. Sequence and Series 8.1 Introduction to Sequence and Its General Terms 8.2 Introduction to Series 8.3 Arithmetic Sequence and Series 8.4 Geometric Sequence and Series 9. Factorization 9.1 Factorization of Expressions in Form of (a ± b)3 and a3 ± b3 9.2 Factorization of Expressions in Form of a4 + a2 b2 + b4 10. HCF and LCM 10.1 Highest Common Factor (HCF) 10.2 Lowest Common Multiple (LCM) 11. Linear Equations 11.1 Methods of Solving Simultaneous Equations 11.2 Verbal Problems on Simultaneous Equations 12. Indices 12.1 Simplification of Indices Chapters / Lessons ALGEBRA UNIT 32 (Th. + Pr.) HSS Way: Mathematics begins at Home, grows in the Surroundings and takes shape in School. IV Specification Grid Unit Areas Total working hour Knowledge Understanding Application Higher ability Total number of items Total number of questions Total Marks No. of items Marks No. of items Marks No. of item Marks No. of item Marks IV Algebra 32 2 2 2 4 3 7 1 2 8 3 15 PB Allied The Leading Mathematics-9 Sequence and Series 201

Algebra Algebra CHAPTER 8 SEQUENCE AND SERIES 8.1 Introduction to Sequence and Its General Terms At the end of this topic, the students will be able to: ¾ find the general or nth term of the linear, quadratic and geometric sequences. Learning Objectives Let's consider the sequences of figures: (i) (ii) Can you add one more figure in the above sequential patterns? Can you find the number of dots in the sixth figure on both sequences? But can you find the number of dots in the 25th figure on both sequences? Can it be done easily? Are they possible? It is so difficult. For these, first we find the rule of the sequence and then find them easily. The rule of the sequence is the specific term for nth term. Let t1, t2, t3, t4, .... tn, ..... be a sequence, then its nth term is denoted by tn. Thus, the standard form of sequence is as t1, t2, t3, t4, .... tn, ..... or, a1, a2, a3, a4, ...........an, ....... or, u1, u2, u3, u3, ........... un, ...... . The nth term of a sequence is known as a general term of the sequence. It is denoted by tn, an or un, etc. and is obtained by the rule, which describes the specific order of the terms of the sequence. A sequence is the collection of objects or numbers in specific rule. Types of Sequence There are following types of sequences. (i) Finite and Infinite Sequences: A sequence having countable number of terms is called a finite sequence. A sequence having uncountable number of terms is called an infinite sequence. For example; 1, 3, 5, 7 is a finite sequence and 3, 6, 9, 12, .... is an infinite sequence. (ii) Linear and Quadratic Sequences: A sequence having constant common difference at the first step of successive terms is called a linear sequence. A sequence having constant common difference at the second step of successive terms is called a quadratic sequence. For example; 2, 4, 6, 8 is a linear sequence and 3, 5, 8, 12, 17, .... is a quadratic sequence. (iii) Arithmetic and Geometric Sequences: A sequence having constant common difference of any two successive terms is called an arithmetic sequence. A sequence having constant common ratio of any two successive terms is called a geometric sequence. For example; 2, 6, 10, 14, ..... is an arithmetic sequence and 1, 2, 4, 8, 16, .... is a geometric sequence. , , , , ...... , ....... , , , , ...... , ..... 202 Allied The Leading Mathematics-9 Sequence and Series 203

Algebra Algebra General Term of Linear or Arithmetic Sequence Consider a linear sequence, 2, 9, 16, 23, ............ 1st Difference: 7 7 7 ........... The linear sequence has constant first differences, that is, the differences of any two consecutive terms are same as at the first step. So, it has the general term tn = an + b, where a, b ∈ Q and n ∈ N. It is the first degree algebraic expression in the variable n, n ∈ N. There are two methods of finding the general term of the linear sequence. They are (i) Method of making pattern (ii) Method of using formula Method-1: Method of making pattern Consider a linear sequence, 2, 9, 16, 23, ............ 1st Difference: 7 7 7 ........... We make a pattern to generalize the given sequence as shown in the table below: Since the first difference is 7. So, what to do in 7, which is equal to 2 ? 7 – 5 = 2 and so on. We have, 7, 14, 21, 28, .... are the multiples of 7. So, take 7 as common in each multiple. Terms 1st Step 2nd Step t1 = 2 7 – 5 7 × 1 – 5 t2 = 9 14 – 5 7 × 2 – 5 t3 = 16 21 – 5 7 × 3 – 5 t4 = 23 28 – 5 7 × 4 – 5 ........ ........ ......... tn ........ 7n – 5 Hence, the general term (or nth term) of the given sequence is 7n – 5, which is also the rule of that sequence. Also, tn = 7n – 5, in which d = 7 and (a – d) = –5. So, we formulate the nth term for the linear sequence as tn = dn + (a – d) = dn + a – d = a + (n – 1)d. Method-2: Method of using formula The given sequence is 2, 9, 16, 23, ..... Here, the first term (a) = 2 The constant or common difference (d) = 9 – 2 = 7 Now, we have The nth term (tn) = dn + (a – d) = 7n + (2 – 7) = 7n – 5 Thus, the general term or the nth term of the given sequence is 7n – 5. Alternatively The nth term (tn) = a + (n – 1)d = 2 + (n – 1)7 = 2 + 7n – 7 = 7n – 5. 202 Allied The Leading Mathematics-9 Sequence and Series 203

Algebra Algebra General Term of Quadratic Sequence Consider a quadratic sequence 2, 5, 10, 17, ... . The above sequence has not equal difference at first, but equal difference at second. This type of sequence is so called the quadratic sequence. Thus, its general term has second degree of algebraic expression as the form of an2 + bn + c, where a, b, c ∈ Q and n ∈ N. The finding of the general term of these types of sequences by the following methods: Method-1: By making pattern Consider a quadratic sequence, 2, 5, 10, 17, 26, ... 1st Difference: 3 5 7 9 ... 2nd Difference: 2 2 2 ... We make a pattern to generalise the above sequence as shown in the table below: Terms 1st Step 2nd Step t1 = 2 1 + 1 12 + 1 t2 = 5 4 + 1 22 + 1 t3 = 10 9 + 1 32 + 1 t4 = 17 16 + 1 42 + 1 t5 = 26 25 + 1 52 + 1 ... ... ... tn ... n2 + 1 Rough We first know, tn = an2 + ..... The second difference, 2a = 2 ∴ a = 1 ∴ tn = 1 × n2 + .... = n2 + ..., where n∈N. So, which is added to the sequence of natural number equals to 2 and so on ? Since 1, 4, 9, 16, 25 ... are the square of natural numbers. So, square each of them. Hence, the general term (or nth term) of the given sequence is n2 + 1, which is also the rule of that sequence. Method-2: By using formula: The given sequence is 2, 5, 10, 17, 26, ... 1st Difference: 3 5 7 9 ... 2nd Difference: 2 2 2 ... Now, we have The general term of the given quadratic sequence is tn = an2 + bn + c ............... (i) Second difference, 2a = 2 or, a = 1 2, 5, 10, 17, 26, ... 1st Difference: 3 5 7 9 ... 2nd Difference: 2 2 2 ... ALGEBRA ALGEBRA 40 Illustrated Optional Maths-9 Thus, its general term has second degree of algebraic expression as the form of an2 + bn + c, where a, b, c ∈ Q and n ∈ N. The finding of the general term of these types of sequences by the following methods: Method-1: By making pattern Consider a quadratic sequence, 2, 5, 10, 17, 26, ... 1st Difference: 3 5 7 9 ... 2nd Difference: 2 2 2 ... We make a pattern to generalise the above sequence shown in the table below: Terms 1st Step 2nd Step t1 = 2 1 + 1 12 + 1 t2 = 5 4 + 1 22 + 1 t3 = 10 9 + 1 32 + 1 t4 = 17 16 + 1 42 + 1 t5 = 26 25 + 1 52 + 1 ... ... ... tn ... n2 + 1 Rough We first know, tn = an2 + ..... The second difference, 2a = 2 ∴ a = 1 ∴ tn = 1 × n2 + .... = n2 + ..., where n∈N. So, which is added to the sequence of natural number equals to 2 and so on ? Since 1, 4, 9, 16, 25 ... are the square of natural numbers. So, square each of them. Hence, the general term (or nth term) of the given sequence is n2 + 1, which is also the rule of that sequence. Method-2: By using formula: The given sequence is 2, 5, 10, 17, 26, ... 1st Difference: 3 5 7 9 ... 2nd Difference: 2 2 2 ... Now, we have The general term of the given quadratic sequence is tn = an2 + bn + c ............... (i) Second difference, 2a = 2 or, a = 1 Substituting the value of a in eqn (i), we get tn = n2 + bn + c ............................ (ii) If n = 1, then t1 = 12 + b × 1 + c or, 2 = 1 + b + c or, b + c = 1 ................................ (iii) If n = 2, then, t2 = 22 + b × 2 + c or, 5 = 4 + 2b + c or, 2b + c = 1 .............................. (iv) a + b + c 4a + 2b + c 9a + 3b + c 16a + 4b + c 3a + b 2a 2a 1 5a + b 7a + b st Difference: 2nd Difference: General form of quadratic sequence is an2 + bn + c. If n = 1, 2, 3, 4, .... then Hence, the second difference of the quadratic sequence is always 2a. 204 Allied The Leading Mathematics-9 Sequence and Series 205

Algebra Algebra Substituting the value of a in eqn (i), we get tn = n2 + bn + c ............................ (ii) If n = 1, then t1 = 12 + b × 1 + c or, 2 = 1 + b + c or, b + c = 1 ................................ (iii) If n = 2, then, t2 = 22 + b × 2 + c or, 5 = 4 + 2b + c or, 2b + c = 1 .............................. (iv) Subtracting the eqn (iii) from eqn (iv), we have 2b + c = 1 b + c = 1 b = 0 – – – Putting the value of b in eqn (iii), we get, 0 + c = 1 or, c = 1. Also, putting the values of b and c in eqn (ii), we get tn = n2 + 0 × n + 1 = n2 + 1, which is the required general term (or nth term). Example-1 Observe the sequence 25, 12, –1, –14, ...... (a) What is the general term of a sequence? (b) Find the nth term of the given sequences. (c) Evaluate the values of 25th term and 50th term of the same sequence. (d) Why is the value of 25th term greater than the value 50th term of the sequence ? Solution: (a) The specific term of a sequence is called its general term and it is also called nth term (tn). (b) Given, 25, 12, –1, 14, ............ 1st Difference: –13 –13 –13 ........... Making a table for arranging the pattern: Terms 1st Step 2nd Step t1 = 25 38 – 13 38 – 13 × 1 t2 = 12 38 – 26 38 – 13 × 2 t3 = –1 38 – 39 38 – 13 × 3 t4 = –14 38 – 52 38 – 13 × 4 ........ ........ ......... tn 38 – 13n (c) The 25th term (t25) = 38 – 13 × 25 = – 287 and the 50th term (t50) = 38 – 13 × 50 = – 612. (d) The value of 25th term is greater than the value 50th term of the sequence becase the sequence is in decreasing order. What to do in –13, it comes 25 ? 204 Allied The Leading Mathematics-9 Sequence and Series 205

Algebra Algebra Alternatively Given, Here, the first term (a) = 25 The common difference (d) = –13 (b) The nth term (tn) = dn + (a – d) = –13n + (25 – (–13)) = –13n + 38 = 38 – 13n (c) The 25th term (t25) = 38 – 13 × 25 = –287 and The 50th term (t50) = 38 – 13 × 50 = – 612. Example-2 Study the sequence 1, 3, 9, 19, 33, ... and write the answers of the given questions. (a) What is the general term of the sequence 2, 4, 6, 8, 10, .... ? (b) Find the nth term of the given sequences. (c) Find the values of the 24th term and 36th term of the sequence. Solution : (a) The general term of the sequence 2, 4, 6, 8, 10, .... is 2n. (b) Given, 1, 3, 9, 19, 33,............ 1st Difference: 2 6 10 14 ....... 2nd Difference: 4 4 4 Method – 1: Making a table for arranging the pattern Terms 1st Step 2nd Step 3rd Step t1 = 1 2 × 12 – 1 2 × 12 – (4 – 3) 2 × 12 – (4 × 1 – 3) t2 = 3 2 × 22 – 5 2 × 22 – (8 – 3) 2 × 22 – (4 × 2 – 3) t3 = 9 2 × 32 – 9 2 × 32 – (12 – 3) 2 × 32 – (4 × 3 – 3) t4 = 19 2 × 42 – 13 2 × 42 – (16 – 3) 2 × 42 – (4 × 4 – 3) t5 = 33 2 × 52 – 17 2 × 52 – (20 – 3) 2 × 52 – (4 × 5 – 3) ........ ........ ........ ........ tn 2n2 – (4n – 3) = 2n2 – 4n + 3 Rough 2nd Difference: 2a = 4 , or, a = 2 ∴ tn = 2n2 + ..... = 2 × 12 + ... 2 × 22 + ...... 2 × 32 + ...... For 1, 5, 9, 13, .... d = 4 ∴tn = 4n – 3. Method-2: By using formula: Let the nth term of the given quadratic sequence be tn = an2 + bn + c ......... (i) Then, the second difference, 2a = 4 or, a = 2 Put a = 2 in eqn (i), we obtain tn = 2n2 + bn + c .......... (ii) 206 Allied The Leading Mathematics-9 Sequence and Series 207

Algebra Algebra If n = 1, then t1 = 2 × 12 + b × 1 + c or, 1 = 2 + b + c or, b + c = –1 ......... (iii) If n = 2, then t2 = 2 × 22 + b × 2 + c or, 3 = 8 + 2b + c or, 2b + c = –5 ........ (iv) Subtracting the eqn (iii) from eqn (iv), we have, 2b + c = –5 b + c = –1 b = – 4 – – + Put b = –4 in eqn (iii), we get – 4 + c = –1 ∴ c = 3 Also, put b = – 4 and c = 3 in eqn (ii), we have tn = 2n2 + (– 4)n + 3 = 2n2 – 4n + 3 (c) Thus, the 24th term (t24) = 2 × 242 – 4 × 24 + 3 = 1059 and The 36th term (t36) = 2 × 362 – 4 × 36 + 3 = 2451 Example-3 Observe the nth term (–1)n (3n – 2) 4n – 1 of a sequence. (a) Write down the first five terms from the given nth term. (b) Find the 30th term of the given sequence. Solution: Given, tn = (–1)n (3n – 2) 4n – 1 Put n = 1, the first term (t1) = (–1)1 (3 × 1 – 2) 4 × 1 – 1 = – 1 3 Put n = 2, the second term (t2) = (–1)2 (3 × 2 – 2) 4 × 2 – 1 = 4 7 Put n = 3, the third term (t3) = (–1)3 (3 × 3 – 2) 4 × 3 – 1 = – 7 11 Put n = 4, the fourth term (t4) = (–1)4 (3 × 4 – 2) 4 × 4 – 1 = 10 15 Put n = 5, the fifth term (t5) = (–1)5 (3 × 5 – 2) 4 × 5 – 1 = – 13 19 Thus, the required first five terms are – 1 3 , 4 7 , – 7 11, 10 15 and – 13 19. Again, the 30th term (t30) = (–1)30 (3 × 30 – 2) 4 × 30 – 1 = 88 119 . 206 Allied The Leading Mathematics-9 Sequence and Series 207

Algebra Algebra PRACTICE 8.1 Keeping Skill Sharp 1. (a) Write the definition of sequence. (b) Write the formula to compute the nth term of an arithmetic sequence. (c) What is the standard form of quadratic sequence? 2. Circle the correct answer. (a) A sequence is a finite sequence if it has ............ terms. (i) infinite (ii) finite (iii) uncountable (iv) complete (b) Which is the general form of a quadratic sequence ? (i) an + b (ii) an + b (iii) 2an + bn + c (iv) an2 + bn + c (c) What is the next term of the sequence 2, 5, 8, 11, 14, ......... (i) 3 (ii) 17 (iii) 20 (iv) 16 (d) What is the next term of the sequence 1, 3, 6, 10, 15, ......... (i) 18 (ii) 19 (iii) 20 (iv) 21 (e) What is the 3rd term of the sequence having nth term tn = 3n - 4? (i) 9 (ii) - 1 (iii) 5 (iv) 4 Check Your Performance 3. Observe the following general term of the sequence and answer the following questions. (a) tn = 2n + 5 (b) tn = 2 – 7n (c) an = (n + 1) (2n – 1) (d) un = (3n – 1)2 (e) tn = 2n2 + 4 (f) an = 1 – 2n 3n + 1 (g) un = (–1)n + 1 (2n2 – 1) (h) tn = (–1)n (3n + 1) 2n (i) un = (–1)n (n2 + 1) (i) What is a sequence? Define it. (ii) Write down the first five terms of the sequences. (iii) Find the difference of the term from each successive term. 4. Study the patterns of the terms of each sequence and answer the following questions. (a) 8, 11, 14, 17 .... (b) 27, 29, 31, 33, .... (c) 50, 45, 40, 35, .... (i) What is the general term of a sequence? Define it. (ii) Find the general term of the given sequences. (iii) Find the 15th term of each sequence. 5. Observe the following sequences and write the answers of the questions. (a) 4, 7, 12, 19, .... (b) 1, 7, 17, 31, .... (c) 2, 1, –2, –7, ... (i) What is an arithmetic sequence? (ii) Add next three terms in each sequence. (iii) Find the general term of the given sequences. 208 Allied The Leading Mathematics-9 Sequence and Series 209

Algebra Algebra 6. Study the following sequences and write the answers of the questions. (a) 1, 2, 3, 4, .... (b) 8, 3, – 2, –7, .... (c) 5, 11, 19, 29, ... (d) 3 5 , 4 7 , 6 9 , 9 11 , ... (e) 0.28, 0.35, 0.42, 0.49, ... (f) 3 2048 , 3 1024 , 3 512 , 3 256 , ... (i) Write any one difference between finite and infinite sequences. (ii) Find the nth term of the given sequences. (iii) Evaluate the 30th term and 45th term. 7. Observe the following patterns of the sequences and answer the questions. (a) (i) What is a quadratic sequence? Define it. (ii) Add the next one figure on each sequence. (iii) Find the nth term of the given sequence. (iv) Find the number of triangles in the 25th figure. (b) (i) Draw the next one shape on the sequence. (ii) Find the nth term of the given sequence. (iii) Find the number of glasses in the 20th shape. (c) (i) Construct next one shape on the sequence. (ii) Find the nth term of the given sequences. (iii) Find the number of vertices in the 30th shape. 3. (a) 7, 9, 11, 13, 15; 2 (b) –5, –12, –19, –26, –33 (c) 2, 9, 20, 35, 54 (d) 4, 25, 64, 121, 196 (e) 6, 12, 22, 36, 54 (f) – 1 4 , – 3 7 , – 5 10 , – 7 13 , – 9 16 (g) 1, –7, 17, –31, 49 (h) –4 2 , 7 4 , –10 6 , 13 8 , –16 10 (i) –2, 5, –10, 17, –26 4. (a) 3n + 5, 50 (b) 2n + 25, 55 (c) 55 – 5n, –20 5. (a) 28, 39, 52; n2 + 3 (b) 49, 71, 97; 2n2 – 1 (c) –14, –23, –34; 1 + 2n – n2 6. (a) tn = n, t30 = 30, t45 = 45 (b) tn = 13 – 5n, t30 = –137 t45 = –212 (c) tn = n2 + 3n + 1, t30 = 991, t45 = 2161 (d) tn = 1 2 n2 – 1 2 n + 3 3 + 2n , t30 = 438 63 , t45 = 993 93 (e) tn = 0.07n + 0.21, t30 = 2.31, t45 = 3.36 (f) 3 × 2n–12, t30 = 786432, t45 = 25769803776 7. (a) 3n – 2; 73 (b) 1 2 (n2 + n); 210 (c) 2n – 1; 536870912 Answers , , , , .... , , , , .... , , , , .... 208 Allied The Leading Mathematics-9 Sequence and Series 209

Algebra Algebra 8.2 Introduction to Series At the end of this topic, the students will be able to: ¾ define series and its types. ¾ solve the problems related to partial sum. Learning Objectives Series Aarisha saves the money everyday adding Rs. 5 more per next day. She starts to save Rs. 10 from the first day for the 7th day of a month. How much money does she save on the seventh day? How much money does she save altogether? To find the answer of the first question, we complete the saving amount per day in the given table: Day 1st 2nd 3rd 4th 5th 6th 7th Amount 10 10 + 5 15 + 5 20 + 5 25 + 5 30 + 5 35 + 5 Sequence 10 15 20 25 30 35 40 Thus, she saves Rs. 40 on the seventh day. But the second question, we need to add all of these amounts of seven days and arrange as a sequence 10, 15, 20, 25, 30, 35, 40. i.e. the sum = 10 + 15 + 20 + 25 + 30 + 35 + 40 = 175. Thus, she saves Rs. 175 altogether for these seven days. The representation of the terms of a sequence in the sum or addition is called a series associated with that sequence. For example, the series of the sequence 2, 6, 14, 26, 42, .... is 2 + 6 + 14 + 26 + 42 + ....... In general, t1, t2, t3, t4, ... , tn, .... is a sequence, then its associated series is represented by Sn and is written as Sn = t1 + t2 + t3 + t4 + .... + tn + .......... Sigma Notation (Using of Summation) Consider a finite series 15 + 21 + 27 + 33 + ... to 10 terms. Now, making pattern Terms 1st Step 2nd Step t1 = 15 6 + 9 6 × 1 + 9 t2 = 21 12 + 9 6 × 2 + 9 t3 = 27 18 + 9 6 × 3 + 9 t4 = 33 24 + 9 6 × 4 + 9 ........ ........ ......... tn ........ 6 × n + 9 We have, the first difference is 6. So, 6 + 9 = 15 and so on. 210 Allied The Leading Mathematics-9 Sequence and Series 211

Algebra Algebra Thus, the sum of the first 10 terms is represented as: S10 = 15 + 21 + 27 + 33 + .... to 10 terms = ∑ (6n + 9) 10 n = 1 If t1 , t2, t3, t4, t5, ..........., tn is a finite sequence, then the series associated with that sequence is t1 + t2 + t3 + t4 + t5 + ........ + tn . The sum of the series is denoted by Sn = t1 + t2 + t3 + t4 + t5 + ........ + tn. and its sigma (summation) notation is symbolized as [t ∑ n] i n = 1 , i ∈ N. where, i = specific term. Here, ∑ is the summation sign. ∑ [tn] i n = 1 is read as "summation of tn 's when n runs from 1 to i". Partial Sum Consider a finite sequence 1, 5, 9, 13, 17, 21, 25, then The sum of the first term, S1 = 1 The sum of the first two terms, S2 = 1 + 5 = 6 The sum of the first three terms, S3 = 1 + 5 + 9 = 15 The sum of the first four terms, S4 = 1 + 5 + 9 + 13 = 28 The sum of the first five terms, S5 = 1 + 5 + 9 + 13 + 17 = 45 The sum of the first six terms, S6 = 1 + 5 + 9 + 13 + 17 + 21 = 66 The sum of the first seven terms, S7 = 1 + 5 + 9 + 13 + 17 + 21 + 25 = 91 Now, taking backward difference every two successive terms, S2 – S1 = 6 – 1 = 5 = t2 (Second term) S3 – S2 = 15 – 6 = 9 = t3 (Third term) S4 – S3 = 28 – 15 = 13 = t4 (Fourth term) ...... S7 – S6 = 91 – 66 = 25 = t7 (Seventh term) Similarly, we generalise for the finite sequence t1, t2, t3, ....... tn–1, tn. Sn – Sn–1 = (t1 + t2 + t3 + ... + tn – 1 + tn) – (t1 + t2 + t3 + ... + tn – 1) = tn. ∴ tn = Sn – Sn – 1 where, n∈N Example-1 Observe the summation or sigma notation 7 n = 3 ∑(4n – 25). (a) What are the values of n in the given summation notation? (b) Write the expanded form of the given summation form. (c) Find its value. Solution: (a) The values of n in the given summation notation are 3, 4, 5, 6 and 7. (b) Here, ∑(4n – 25) 7 n=3 = (4×3 – 25)+(4×4 – 25)+(4×5 – 25)+(4×6 – 25)+(4×7 – 25) = – 13 – 9 – 5 – 1 + 3 = – 25. 210 Allied The Leading Mathematics-9 Sequence and Series 211

Algebra Algebra (c) The value of summation notation is – 25. Example-2 Observe the summation notation 30 n=7 ∑ 3n2 – 21 2n – 21 . (a) How many terms are there in the series as in the form of sigma notation. (b) Find its 15th term. Solution: (a) Here, the given series as in the form of sigma notation is 30 n=7 ∑ 3n2 – 21 2n – 21 There are 30 – 6 = 24 terms in that series and for the 15th term, n = 15 + 6 = 21 (b) The required 15th term (t15) = 3 × 212 – 21 2 × 21 – 21 = 1302 21 = 62. Example-3 Write the sigma notation of the series: 1 + 3 + 7 + 13 + 21 + 31 + 43 Solution: Here, the given series is 1 + 3 + 7 + 13 + 21 + 31 + 43 1st Difference: 2 4 6 8 10 12 2nd Difference: 2 2 2 2 ..... Now, making pattern Terms 1st Step 2nd Step t1 = 1 12 – 0 12 – (1 – 1) t2 = 3 22 – 1 22 – (2 – 1) t3 = 7 32 – 2 32 – (3 – 1) t4 = 13 42 – 3 42 – (4 – 1) ........ ........ ......... t7 = 43 72 – 6 72 – (7 – 1) ........ ........ ......... tn = n2 – (n – 1) = n2 – n + 1 Thus, the required sigma notation is (n2 ∑ – n + 1) 7 n=1 . Example-4 The sum of the first n terms of a sequence is 4n2 – 2n + 1. (a) How many terms are there in the series as in the given sigma notation? (b) Find the 15th term of the sequence. Solution: (a) Here, the sum of the first n terms of the sequence, Sn = 4n2 – 2n + 1 ∴ Sn–1 = 4(n – 1)2 – 2(n – 1) + 1 = 4(n2 – 2n + 1) – 2n + 2 + 1 = 4n2 – 8n + 4 – 2n + 3 = 4n2 – 10n + 7 Now, we have, the nth term, tn = Sn – Sn – 1 = 4n2 – 2n + 1 – (4n2 – 10n + 7) = 4n2 – 2n + 1 – 4n2 + 10n – 7 = 8n – 6 Rough The second difference, 2a = 2 a = 1 ∴ tn = n2 + ... 212 Allied The Leading Mathematics-9 Sequence and Series 213

Algebra Algebra (b) The 15th term, t15 = 8 × 15 – 6 = 114. "Alternatively" Given, Sn = 4n2 – 2n + 1 ∴ S15 = 4 × 152 – 2 × 15 + 1 = 871 and S14 = 4 × 142 – 2 × 14 + 1 = 575. ∴ t15 = S15 – S14 = 871 – 757 = 114. PRACTICE 8.2 Keeping Skill Sharp 1. (a) Define series. (b) What is sigma notation? 2. Represent the following sequences into their associated series: (a) 1, 6, 11, 16, .... to n terms (b) 15, 6, –3, –12, .... to n terms (c) 15, 20, 30, 45, .... to n terms (d) 45, 39, 31, 21, .... to n terms (e) 2, 4, 6, 8, .... to n terms (f) 5, 10, 20, 40, .... to n terms Check Your Performance 3. Observe the nth term of some sequences below where n ∈ N. (a) tn = 2n2 – n + 3 (b) an = 2n + 3 2n – 1 (c) un = (–1)n – 1 3n (d) tn = (–1)n 3n (i) What is nth term ? Define it. (ii) Write the first five terms of each sequence. (iii) Find the sum of the first five terms (S5) of each sequence. 4. Observe the summation or sigma notation and answer the following questions. (a) 4 n=1 ∑ (2n2 – 2n + 3) (b) 6 n=2 ∑ (2n + 1) (3n – 1) (c) 7 n=4 ∑ 2n2 + 3 n – 2 (d) 6 n=2 ∑ (–1)n (2n + 1) (e) 8 n=3 ∑ (–1)n+1 (n2 + 2) (f) 7 n=2 ∑ (– 1)n (2n + 1) 3n – 1 (i) What are the values of n in the given sigma notations? (ii) Write the expanded form of the given summation form. (iii) Find their values. 5. Observe the following sigma notation and answer the following questions. (a) 16 n=2 ∑ xn (b) 25 n=3 ∑ (2n2 + 3) (c) 20 k=5 ∑ (–1)k 4k – 1 k + 1 (i) How many terms are there in the series as in the given sigma notation. (ii) Find its 12th term. 212 Allied The Leading Mathematics-9 Sequence and Series 213

Algebra Algebra 6. Write each of the following by using sigma notation: (a) 4 + 7 + 10 + 13 + 16 + 19 + 22 (b) 2 5 + 6 7 + 10 9 + 14 11 + 18 13 + 22 15 (c) – 3 5 + 5 9 – 7 13 + 9 17 – 11 21 (d) 1 + 5 + 25 + 125 + 625 + 3125 (e) (r – 1) 2 + (r – 1)2 4 + (r – 1)3 8 + (r – 1)4 16 + (r – 1)5 32 + (r – 1)6 64 (f) a + ar + ar2 + ar3 + ar4 + ar5 + ar6 (i) Find the nth term of the given series. (ii) Write each sequence by using sigma notation. 7. Observe the following series whose sum of the first n terms are given below: (a) Sn = 2n + 1 (b) Sn = 3n2 – 1 (c) Sn = (2n – 1)2 (d) Sn = 2n2 + 4n (i) Find the sum of the first four terms from the given sum of n terms. (ii) Find the sum of the first 25 terms. (iii) Find the sum of the first (n – 1) terms from the given sum of the first n terms. (iv) Find the nth term from the given sum of n terms. 2. (a) 1 + 6 + 11 + 16 + .. + (5n – 4) (b) 15 + 6 – 3 – 12 – ... – (24 – 9n) (c) 15 + 20 + 30 + 45 + .. + 5 2 n2 – 5 2 n + 15 (d) 45 + 39 + 31 + 21 + ... + (– n2 – 3n + 49) (e) 2 + 4 + 8 + 16 + .... + (2n ) (f) 5 + 10 + 20 + 40 + .... + (5 × 2n–1) 3. (a) 110 (b) 12 47 315 (c) 183 (d) –183 3. (a) 4, 9, 18, 31, 48; 110 (b) 5, 7 3 , 9 5 , 11 7 , 13 9 ; 12 47 315 (c) 3, –9, 27, –81, 243; 183 (d) –3, 9, – 27, 81, –243; – 183 4. (a) 1, 2, 3, 4; 3 + 7 + 15 + 27 + 43; 95 (b) 2, 3, 4, 5, 6; 25 + 56 + 99 + 154 + 221; 555 (c) 4, 5, 6, 7; 35 2 + 53 3 , 75 4 , 101 5 ; 4447 60 (d) 2, 3, 4, 5, 6; 5 – 7 + 9 – 11 + 13; 9 (e) 3, 4, 5, 6, 7, 8; 11 – 18 + 27 – 38 + 51 – 66; – 33 (f) 2, 3, 4, 5, 6, 7; 1 – 7 8 + 9 11 – 11 14 + 13 17 – 15 20 ; 1803 10472 5. (a) 15, x13 (b) 23, 395 (c) 16, 63 17 6. (a) 3n + 1; 7 n=1 (3n + 1) (b) 4n – 2 2n + 3 ; 6 n=1 4n – 2 2n + 3 (c) (–1)n (2n + 1) 4n + 1 , 5 n=1 (–1)n (2n + 1) 4n + 1 (d) 5n–1, 6 n=1 5n–1 (e) (r – 1)n 2n , 6 n=1 (r – 1)n 2n (f) arn–1, 7 n=1 arn–1 7. (a) 9; 51; 2n – 1; 2 (b) 47; 1874; 3n2 – 6n + 2; 6n – 3 (c) 49; 2401; (2n – 3)2 ; 8n – 8 (d) 48; 1350; 2n2 – 2; 4n + 2 Answers 214 Allied The Leading Mathematics-9 Sequence and Series 215

Algebra Algebra 8.3 General Term of Arithmetic Sequence At the end of this topic, the students will be able to: ¾ find the general term of arithmetic sequence. ¾ solve the problems related to arithmetic sequence. Learning Objectives Arithmetic Progression (Sequence and Series) Let's suppose that the first term is 3 and the common difference is 5 in an AP, it will be; 3, 3 + 5, 8 + 5, 13 + 5, …….. = 3, 8, 13, 18, …… . 3, 8, 13, 18, ....... Difference: 5 5 5 ............. → Common difference A sequence/series having the constant difference between any term and its preceding one is called an arithmetic progression (sequence and series)/AP or AS. The constant difference is called the common difference (d). Generally, we suppose that if t1, t2, t3, t4, …………, tn-1, tn or t1 + t2 + t3 + t + ……… + tn-1 + tn is an AP, then the common difference of AP is d = t2 – t1 = t3 – t2 = …… = tn – tn-1 i.e., The common difference (d) = Any term – its preceding term (except first term) ∴ Any term = preceding term + common difference (except first term) If a is the first term and d is the common difference of an AP, we can generalize the following AP: First Term Second Term Third Term Fourth Term ……… t1 t2 t3 t4 ……… a a + d a + d + d = a + 2d a + 2d + d = a + 3d ……… General Term or nth Term of AP Consider an AP 2, 5, 8, 11, …… . What is its 100th term ? Can you find easily ? Then, first, we find its general term or rule. In the above sequence, First term (t1) = 2 = 3 – 1 = 3 × 1 – 1 Second term (t2) = 5 = 6 – 1 = 3 × 2 – 1 Third term (t3) = 8 = 9 – 1 = 3 × 3 – 1 Fourth term (t4) = 11 = 12 – 1 = 3 × 4 – 1 ……….... …. ...… ….....… The nth term (tn) = 3n – 1. So, the general term of the AP 2, 5, 8, 11, ..… is 3n – 1, which is the rule of the progression. But 214 Allied The Leading Mathematics-9 Sequence and Series 215

Algebra Algebra now, we form a formula for computing the general term (or nth term). If a and d are the first term and the common difference of an AP respectively, we make a table as below: Term Step 1 Step 2 First term (t1) a + 0 × d a + (1 – 1)d Second term (t2) a + 1 × d a + (2 – 1)d Third term (t3) a + 2 × d a + (3 – 1)d Fourth term (t4) a + 3 × d a + (4 – 1)d …… …… …… nth term (tn) …… a + (n – 1)d Thus, the general term of the AP is tn= a + (n – 1)d, where a = first term, d = common difference, n = number of term, tn = nth term. Properties of AP (i) If each term of an AP is increased or decreased by a constant quantity, the resulting quantities are in AP with the same common difference. For example; If a, b, c, d are in AP, then a + x, b + x, c + x, d + x and a – x, b – x, c – x, d – x are also in AP. (ii) If each termof anAPismultiplied or divided by a constant quantity, the resulting quantities are inAPwiththesamecommondifference.Forexample;Ifa,b,c,dareinAP,thenax,bx,cx,dxand a x, b x, c x, d x are also in AP. Alert to Arithmetic Sequence 1. If the difference of any two consecutive terms are not equal, it is not AS or AP. 2. If the difference of any two consecutive terms are equal second times, it is also AP in quadratic form. 3. We do not multiply for finding the forward term to the just preceding term by the common difference. 4. The number of terms never be zero or negative or fractional or irrational number in the sequence. Points to be Noted 1. A sequence in which the difference between two consecutive terms is always a constant, is called an arithmetic sequence. The standard form of AP is a, a + d, a + 2d, a + 3d, a + 4d, ………., a + (n – 1)d. 2. The constant difference of an AP is called the common difference (d). If t1, t2, t3, t4, …………, tn-1, tn is an AP, then the common difference of AP is, d = t2 – t1 = t3 – t2 = …… = tn – tn-1 = Any term – Its preceding term (except first term) 3. The general term or rule of the AP is tn= a + (n – 1)d, where a = first term, d = common difference, n = number of term, tn = nth term. Last term (l or b), tn / l / b = a + (n – 1)d 216 Allied The Leading Mathematics-9 Sequence and Series 217

Algebra Algebra Example-1 The first term and the common difference of an AP are – 5 and 2 respectively. (a) Write the definition of an arithmetic sequence. (b) Find the arithmetic sequence from the given information. Solution: (a) A sequene with constant different of any two successive terms is called an arithmetic sequence. (b) Given, in an AP the first term (a) = – 5, the common difference (d) = 2. Now, The first term (t1) = a = – 5 The second term (t2) = a + d = – 5 + 2 = – 3 The third term (t3) = a + 2d = – 5 + 2 × 2 = – 1 The fourth term (t4) = a + 3d = – 5 + 3 × 2 = 1 and so on. Hence, the required AP is –5, –3, –1, 1, …………. . Example-2 Observe the AS 20, 8, -4, -16, … . (a) Find the general term of the given arithmetic sequence. (b) find its 50th term. Solution: Here, the given arithmetic progression is 20, 8, –4, –16,…….., in which the first term (a) = 20, the common difference (d) = 8 – 20 = – 4 – 8 = –16 – (–4) = –12. Now, we have, (a) The general term (tn) = a + (n – 1)d = 20 + (n – 1)(–12) = 20 – 12n + 12 = 32 – 12n. (b) Again, the number of terms (n) = 50 ∴ The 50th term (t50) = 32 – 12×50 = –568. Example-3 Does –15 belong to the arithmetic series 15 + 121 2 + 10 + 71 2 + …… ? Solution: Given, in the arithmetic series, the first term (a) = 15, common difference (d) = 121 2 –15 = 21 2 the last term (l) = –15 (suppose), the number of term (n) = ? Now, we have , l = a + (n–1)d or, –15 = 15 + (n–1) – 2 1 2 or, –15 – 15= (n–1) – 5 2 or, 30 × 2 5 = n – 1 or, 12 + 1 = n or, n = 13, which is the counting or natural number. So, –15 belongs to the given series. If the number of terms (n) is positive integer, we reach the correct answer. But, n is in negative integer, the given term does not belong to the given sequence. 216 Allied The Leading Mathematics-9 Sequence and Series 217

Algebra Algebra Note: If n is negative (or not a whole number), the given term/number will not be the term of the given sequence. Example-4 Study the first three terms k – 2, k + 3 and 3k + 2 of an AS. (a) Find the value of k. (b) Find these three terms. (c) Compute the next three terms. (d) Find its nth term. Solution: (a) Since k – 2, k + 3 and 3k + 2 are in AP, so (k + 3) – (k – 2) = (3k + 2) – (k + 3) or, k + 3 – k + 2 = 3k + 2 – k – 3 or, 5 = 2k – 1 or, 5 + 1 = 2k or, 2k = 6 ∴ k = 3. (b) The required three terms are 3 – 2, 3 + 3, 3 × 3 + 2 = 1, 6, 11. (c) Here, the common difference (d) = 6 – 1 = 11 – 6 = 5. So, the required next three terms are 11 + 5 = 16, 16 + 5 = 21, 21 + 5 = 26. (d) The required nth term (tn) = a + (n – 1)d = 1 + (n – 1) × 5 = 1 + 5n – 5 = 5n – 4. Example-5 If the fifth term and tenth term of an AP are 17 and 42 respectively, find its 1000th term. (a) Find the common difference of the AS. (b) Find its first term. (c) Find its 1000th term. Solution: (a) Here, the fifth term (t5) = 17 or, a + (5 – 1)d = 17 or, a + 4d = 17 …………. (i) the tenth term (t10) = 42 or, a + (10 – 1)d = 42 or, a + 9d = 42 .................... (ii) Subtracting eqn (i) from eqn (ii), we get a + 9d = 42 a + 4d = 17 5d = 25 ∴ d = 5. (b) For the first term, substituting the value of d in eqn (i), we get a + 4 × 5 = 17 or, a = 17 – 20 ∴ a = – 3. Oh! I know, the backward differences are equal. Oh! I know, tn = a + (n – 1)d – – – 218 Allied The Leading Mathematics-9 Sequence and Series 219

Algebra Algebra (c) The first term (a) = – 3, the common difference (d) = 5 The number of terms (n) = 1000, the 1000th term (t1000) = ? Now, we have t1000 = – 3 + (1000–1)5 = – 3 + 999 × 5 = – 3 + 4995 = 4992. Example-6 Obseve the arithmetic sequence 3, 5, 7, 9, …..., 201. (a) Find the common difference of the AS. (b) Find its number of terms. (c) Find the 12th term from its end. Solution: (a) Here, in an AS 3, 5, 7, 9, ………., 201, First term (a) = 3, Common difference (d) = 5 – 3 = 2, Last term (l) = 201, Number of terms (n) = ? (b) Now, we have, l = a + (n – 1)d or, 201 = 3 + (n – 1) × 2 or, 201 – 3 = (n – 1) × 2 or, 192 2 = n – 1 or, 99 + 1 = n or, n = 100. (c) Then, the 12th term from the end = the [100 – (12 – 1)] = 89th term from the beginning. So, t89 = a + (n – 1)d = 3 + (89 – 1) × 2 = 3 + 88 × 2 = 179. Hence, the required 12th term from the end of the given AS is 179. “Alternatively” Here, in an AS 3, 5, 7, 9, ……….201. The first term from the end (a) = 201 The common difference from beginning = 5 – 3 = 2 ∴ The common difference from end (d) = – 2 The number of terms from the end (n) = 12 The 12th term from the end (t12) = ? Now, we have t12 = a + (n – 1)d = 201 + (12 – 1) (–2) = 201 – 22 = 179. Hence, the required 12th term from the end of the given AS is 179. Example-7 The 6th term of an AP is 5 times the first term and the 11th term exceeds double the 5th term by 3. (a) Find the common difference of the AS. (b) Find its first term. (c) Find its 20th term. Solution: (a) Here, in an AP, the 6th term (t6) = 5 × the first term (a) 218 Allied The Leading Mathematics-9 Sequence and Series 219

Algebra Algebra or, a + 5d = 5a [ tn = a + (n–1)d] or, 4a – 5d = 0 ………………. (i) Again, the 11th term (t11) = 2 × the 5th term (t5) + 3 or, a + 10d = 2(a + 4d) + 3 or, a + 10d = 2a + 8d + 3 or, 2a – a = 10d – 8d – 3 or, a = 2b – 3 ................. (ii) Putting the value of a from eqn (ii) in eqn (i), 4(2d – 3) –5d = 0 or, 8d – 12 – 5d = 0 or, 3d – 12 = 0 or, 3d = 12 or, d = 4. (b) Putting the value of d in eqn (ii), we get a = 2 × 4 – 3 = 8 – 3 = 5 (c) Now, we have tn = a + (n – 1)d, The 20th term (t20) = 5 + (20 – 1)4 = 5 + 19×4 = 5 + 76 = 81. Example-8 The sum of three consecutive numbers in AP is 27 and the sum of the square of the first and last terms is 194. (a) Suppose the three consecuitive numbers in AP. (b) Find these three umbers. Solution: Let, a – d, a, a + d be three consecutive numbers in AP, then a – d + a + a + d = 27. or, 3a = 27 ∴ a = 9. Again, (a – d)2 + (a + d)2 = 194 or, (9 – d)2 + (9 + d)2 = 194 or, 81 – 18d + d2 + 81 + 18d + d2 = 194 or, 2d2 = 194 – 162 or, 2d2 = 32 or, d2 = 16 ∴ d = ± 4. If a = 9 and d = 4 then the required three numbers are 9 – 4, 9, 9 + 4 = 5, 9, 13. If a = 9 and d = – 4 then the required three numbers are 9 – (–4), 9, 9 + (–4) = 13, 9, 5. 220 Allied The Leading Mathematics-9 Sequence and Series 221

Algebra Algebra Example-9 The lengths of sides of a quadrilateral are in AS. If the perimeter of the quadrilateral is 28 cm and the length of the longest side is 10 cm, find the length of other sides. Solution: Let a – 3d, a – d, a + d and a + 3d be the length of four sides of the quadrilateral in AS. Then the perimeter (P) = 28 or, a – 3d + a – d + a + d + a + 3d = 28 or, 4a = 28 ∴ a = 7. Again, the largest side, a + 3d = 10 or, 7 + 3d = 10 or, 3d = 10 – 7 or, 3d = 3 ∴ d = 1. Hence, the required lengths of the sides of the quadrilateral are; 7 – 3 × 1, 2 – 1, 7 + 1 = 4 cm, 6 cm, 8 cm. Note: Table of Assumption of Terms in AP SN Property Terms in AP 1. Three consecutive/successive terms/ numbers in AP. a – d, a, a + d 2. Five consecutive terms in AP. a – 2d, a – d, a, a + d, a + 2d 3. Four consecutive terms in AP. a – 3d, a – d, a + d, a + 3d 4. Six consecutive terms in AP. a – 5d, a – 3d, a – d, a + d, a + 3d, a + 5d Example-10 A private company gives Rs. 20000 per month for a job of officer at the beginning and increases Rs. 1800 in the salary per year as experience. What is the salary of the officer during the tenth year? Solution: Here, the initial salary (a) = Rs. 20000, the increasing salary per year (d) = Rs. 1800 the duration of time (in year) (n) = 10, the salary at the 10th year (tn) = ? Now, we have, t10 = 20000 + (10 – 1)1800 [ tn = a + (n – 1)d] = 20000 + 9 × 1800 = 20000 + 16200 = 36200. Hence, the salary of the officer is Rs. 36200 during the tenth year. 220 Allied The Leading Mathematics-9 Sequence and Series 221

Algebra Algebra PRACTICE 8.3 Read Think Understand Do Keeping Skill Sharp 1. (a) Define arithmetic sequence. (b) In which condition does a sequence become arithmetic? (c) What is the common difference in the sequence a + d, a, a – d, a – 2d, .............? 2. (a) If the first term and the common difference of an AS are 7 and –3 respectively then find the first five terms of the AS. (b) If the first term and the common difference of arithmetic sequence are 2 3 and 1 3 respectively then find the arithmetic sequence. Check Your Performance 3. Observe the following arithmetic sequences: (i) 101, 120, 139, 158, ...... (ii) 120 + 114 + 108 + 102 + ...... (a) Write the formula to find the nth term of the arithmetic sequence. (b) Find the common difference of the arithmetic sequence. (c) Find the nth term of the arithmetic sequence. (d) Find its 20th term. 4. (i) If the 10th term of an AP having the common difference 5 is 43, find the first term. (ii) The 12th term of an arithmetic series, whose first term is 0.35, is 0.02. Find its common difference. 5. (i) How many terms are there in the AS 2, 9, 16, 23, …………., 142 ? (ii) Which term of the arithmetic series 21 + 25 + 29 + 33 + ……. is 397? (iii) Is – 29 3 a term of the series 19 3 + 13 3 + 7 3 + ……… ? 6. (i) Study the first three terms 2a + 1, 3a, 5a + 1 of an AP. (ii) Study the first three terms a, a2 + 1 and a + 6 of an AS. (a) Find the value of a. (b) Find these three terms. (c) Compute the next three terms. (d) Find its nth term. 7. (i) The sixth and seventeenth terms of an arithmetic progression are 19 and 41 respectively. (a) Find the common difference of the AS. (b) Find its first term. (c) Is 36 a term of the sequence? (d) Find its 100th term. 222 Allied The Leading Mathematics-9 Sequence and Series 223

Algebra Algebra (ii) The 5th term and 8th term of an AP are 19 and 31 respectively. (a) Find the common difference of teh AS. (b) Find its first term. (c) Which term of it is 67 ? (d) Find its 80th term. 8. (i) 8 times of the eighth term of an AP is equal to 12 times of the twelfth term. Find its first term if the common difference is – 2. (ii) 8 times of the third term and double of the tenth term of an AP are equal to each other. Find the common difference of the AP if its first term is 2. 9. (i) The 5th term of an AP with first term 1 40 is 1 8 . (a) Find the common difference of the AS. (b) Find its 11th term. (ii) The third term and the sixth term of an AP are 1 and 2 respectively. (a) Find the common difference of the AS. (b) Find its first term. (c) Find its 27th term. (iii) If 5 times of the fifth term is equal to 12 times of the twelfth term of an AP, find its 17th term. 10. (i) The sum of three successive numbers in an AP is 54 and the product of the last two terms is 333. (ii) Divide 36 into three parts which are in AP such that their product is 756. (a) Suppose the three consecutive numbers in AP. (b) Find these three numbers. 11. (i) A man is appointed on a salary of Rs. 15500 per month. He gets an increment of Rs. 1500 as grade every year. What is his monthly salary in the fifth year? (ii) The meter of a taxi starts with a reading of Rs. 15 and then runs up Rs. 14 for each kilometer travelled. How much should be paid for a journey of 12 km? 2. (a) 7, 4, 1, –2, –5 (b) 2 3 , 1, 4 3 , 5 3 , 2 3. (i) 19, 19n + 82, 462 (ii) – 6, 126 – 6n, 6 4. (i) – 2 (ii) – 0.03 5.(i) 22 (ii) 95 (iii) Yes 6. (i) (a) – 2 (b) –3, –6, –9 (c) –12, –15, –18 (d) –3n (ii) (a) If a = 2, (b) 2, 5, 8 (c) 11, 14, 17 (d) 3n – 1 If a = –1, (b) –1, 2, 5 (c) 8, 11, 14 (d) 3n – 4 7. (i) (a) 2 (b) 9 (c) No (d) 207 (ii) (a) 4 (b) 3 (c) 17th (d) 319 8. (i) 8 (ii) 6 9.(i) (a) 1 40 (b) 11 40 (ii) (a) 1 3 (b) 1 3 (c) 9 10. (i) (b) 17 7 8 , 18, 18 7 8 (ii) (b) 3, 12, 21, or 21, 12, 3 11. (i) Rs. 2150 (ii) Rs. 183 Answers 222 Allied The Leading Mathematics-9 Sequence and Series 223

Algebra Algebra 8.4 General Term of Geometric Sequence At the end of this topic, the students will be able to: ¾ find the general term of geometric sequence. ¾ solve the problems related to geometric sequence. Learning Objectives Introduction to Geometric Progression Consider the sequence as follows: SN Sequences Common Ratios (i) 1, 2, 4, 8, …........… r = 2 1 = 4 2 = 8 4 = ……....….. = 2 (ii) 16 + 8 + 4 + 2 + …. r = 8 16 = 4 8 = 2 4 = ……….... = 1 2 (iii) 1, 1 3, 1 9, 1 27, …....… r = 1/3 1 = 1/9 1/3 = 1/27 1/9 = ….. = 1 3 (iv) a, ar, ar2 , ar3 , …...... r = ar a = ar2 ar = ar3 ar2 = …..….. = r These sequences have the same (or constant) backward ratio that are called geometric progression. A sequence or series having the constant/same ratio between any term and its preceding one is called a geometric progression. The constant ratio is called the common ratio (r). Generally, we suppose that if t1, t2, t3, t4……, tn-1, tn or, t1 + t2 + t3 + t4 + ….. + tn-1 + tn is an GP, then the common ratio of GP is given by, r = t2 t1 = t3 t2 = t4 t3 = ........... = tn tn – 1 . i.e., Common ratio (r) = Any term Its preceding term If a and r are the first term and common ratio of an GP respectively, then we can generalize the following GP: First Term Second Term Third Term Fourth Term …………… t1 t2 t3 t4 ………….... a ar ar.r = ar2 ar2 .r = ar3 …………… For example: If 4 is the first term and 2 is the common ratio then the GP will be 4, 4 × 2, 8 × 2, 16 × 2, …………….. = 4, 8, 16, 32, ……… . 1 224 Allied The Leading Mathematics-9 Sequence and Series 225

Algebra Algebra General Term of GP Consider a GP 1, 2, 4, 8, 16, …….., . What is its 200th term ? Can you find easily ? For this, at first, we find its rule, called general term. In the above sequence, The first term (t1) = 1 = 20 = 21-1 The second term (t2) = 2 = 21 = 22-1 The third term (t3) = 4 = 22 = 23-1 The fourth term (t4) = 8 = 23 = 24-1 ………………… ……..…. . The nth term (tn) = 2n-1 So, the general term of the GP 1, 2, 4, 8, …….. is 2n – 1, which is the rule of the progression. But, now we form a formula for computing the general term (or nth term) as below. If a and r are the first term and the common ratio of GP respectively, then we make the table as given below: Terms Step 1 Step 2 First Term (t1) = a.r0 = ar1–1 Second Term (t2) = a.r1 = ar2–1 Third Term (t3) = a.r2 = ar3–1 Fourth Term (t4) = a.r3 = ar4–1 ………… ………… …………. nth Term (tn) ………… = arn–1 Thus, the general term of the GP is tn = arn-1, Where, a = first term, r = common ratio n = number of term, tn = nth term. Note: The last term (l or b) is represented by the nth term (tn). So, tn or l or b = arn-1. Properties of GP (i) If all the terms of a GP are multiplied or divided by the same quantity, then the product (or quotient) will be in GP. (ii) The reciprocals of the terms of a GP are also in GP. (iii) If each term of a GP is raised to the same power, the resulting terms will also form a GP. If a, b, c, d, ...... are in GP, then (a) 1 a, 1 b, 1 c, 1 d …….. are in GP. (b) am, bm, cm, dm, ……... are in GP. (iv) ax, bx, cx, dx, …… are in GP, and a x, b x, c x, d x, …….. are in GP. Newly-infected people Stay safe from COVID-19 224 Allied The Leading Mathematics-9 Sequence and Series 225

Algebra Algebra Alert to Geometric Sequence 1. If the ratio of any two consecutive terms is not equal, it is not GS or GP. 2. We do not add for finding the forward term to the just preceding term by the common ratio. Points to be Noted 1. A sequence in which the ratio of two consecutive terms is always a constant, is called an geometric sequence. The standard form of AP is a, ar, ar2 , ar3 , ar4 , ………., arn-1. 2. The constant ratio of a GP is called the common ratio (r). If t1, t2, t3, t4, …………, tn-1, tn is a GP, then the common ratio of GP is r = t2 t1 = t3 t2 = ........ = tn tn – 1 = Any term Its preceding term . 3. The general/nth term or rule of the GP is tn= arn-1, where a = 1st term, r = common ratio, n = No. of term. Example-1 Observe the following sequences. Which are geometric sequences? (i) 8, 24, 72, 216, ….….. (ii) 8, 24, 40, 56, ……… Solution: (i) Here, the given sequence is 8, 24, 72, 216, …….. Backward Ratio: 3 3 3 …….. Since the backward ratios of any two consecutive terms are constant, so the given sequence is a geometric sequence. (ii) Here, the given sequence is 8, 24, 40, 56, ……… Backward Ratio: 3 5 3 7 5 ……… Since the backward ratios of any two successive terms are not constant, so the given sequence is not a geometric sequence. Example-2 If the first term and the common ratio of a GP are 2 and 3 respectively, find the GP. (a) Write the definition of a geometric sequence. (b) Find the geometric sequence from the given information. Solution: (a) A sequence having constant ratio of any two successive terms is called a geometric sequence (b) Given, the first term (a) = 2, the common ratio (r) = 3 Now, we have, the first term (t1) = a = 2, Second term (t2) = ar = 2 × 3 = 6, Third term (t3) = ar2 = 2 × 32 = 18, 226 Allied The Leading Mathematics-9 Sequence and Series 227

Algebra Algebra Fourth term (t4) = ar3 = 2 × 33 = 54 and so on. Hence, the required GP is 2, 6, 18, 54, …………..……. . Example-3 Observe the geometric sequence 1, 4, 16, 64, ........... . (a) Write the formula to find the general term of a geometric sequence. (b) Compute the common ratio of the given geometric sequence. (c) Find the general term (nth term) of the given geometric sequence. (d) Find its 10th term. Solution: (a) General term of a geometric sequence (tn) = arn – 1 (b) Here, in the given GP is 1, 4, 16 , 64, .........., First term (a) = 1, Common ratio (r) = 4 1 = 4. (c) Now, we have The general term (tn) = arn–1 = 1.4n–1 = 4n–1. (d) The 10th term (t10) = 410–1 = 49 = 262144. Example-4 Does 1 2048 belong to the geometric series 128 + 64 + 32 + 16 + ……? Solution: Here, the given geometric series is 128 + 64 +32 + 16 + ……… in which, First term (a) = 128, Common ratio (r) = 64 128 = 1 2, Last term (l) = 1 2048 Number of terms (n) = ? Now, we have, l = arn – 1 or, 1 2048 = 128. 1 2 n – 1 or, 1 2048 × 128 = 1 2 n – 1 or, 1 2 n – 1 = 1 2 18 ∴ n – 1 = 18 or, n = 18 + 1 = 19, which is counting number. Yes, 1 2048 belongs to the given series. Example-5 p + 4, p – 2 and p – 6 are three successive terms in a GP. (a) Find the value of p. (b) Find these three terms. (c) Compute the next three terms. (d) Find its nth term. Solution: (a) Since p + 4, p – 2 and p – 6 are three terms in GP. So, p – 2 p + 4 = p – 6 p – 2 or, p2 – 4p + 4 = p2 – 2p – 24 or, 2p = 28 => p = 14 Hence, these three terms are 14 + 4, 14 – 2, 14 – 6 = 18, 12, 8. 226 Allied The Leading Mathematics-9 Sequence and Series 227

Algebra Algebra (b) The required three terms are 14 + 4, 14 – 2 and 14 – 6 = 18, 12, 8. (c) Here, the common ratio (r) = 12 18 = 8 12 = 2 3 ∴ The required next three terms are 8 × 2 3 = 16 3 , 16 3 × 2 3 = 32 3 and 32 3 × 2 3 = 64 3 . (d) The required nth term (tn) = arn – 1 = 18 × ( 2 3) n – 1 = 18(2 3) n – 1 . Example-6 The seventh term and twelfth term of a GP are 8 and 256 respectively. (a) Find its common ratio. (b) Calculate the and first term of the progression. (c) Find its 20th term. Solution: (a) Here, in the given GP, the seventh term (t7) = 8, or, ar7–1 = 8 or, ar6 = 8 ………………... (i) and the twelfth term (t12) = 256 or, ar12–1 = 256 or, ar11 = 256 ………...…… (ii) Dividing the eqn (i) by the eqn (ii), we get ar11 ar6 = 259 8 or, r11–6 = 32 or, r5 = 25 => r = 2 (b) Put r = 2 in eqn (i), we get a.26 = 8 or, a × 64 = 8 or, a = 8 64 = 1 8 (c) The required twentieth term (t20) = arn - 1 = 1 8 × 220 – 1 = 219 8 = 524288 8 = 65536. Example-7 The fifth term of a GP having positive ratio is 4 times of the third term and that of the eighth term is 184. (a) Find its common ratio. (b) Find the and first term of the progression. (c) Find its 15th term. Solution: (a) Here, in the given GP, Fifth term (t5) = 4 × Third term (t3) or, ar4 = 4 × ar2 or, r2 = 4 [ by given, the ratio is positive.] or, r = 2. (b) The eighth term (t8) = 384 ar7 = 384 228 Allied The Leading Mathematics-9 Sequence and Series 229

Algebra Algebra or, a × 27 = 384 or, a = 384 128 = 3. (c) The required fifteenth term (t15) = 3 ×215-1 [ tn = arn – 1] = 3 × 214 = 3 × 16384 = 49152. Example-8 Three consecutive terms are in a GP and its product is 64. If the sum of the first and last term is 10, then find the three terms. (a) Suppose the three consecutive numbers in AP. (b) Find these three numbers. Solution: (a) Let a r , a, ar be three terms in a GP. (b) Then, a r × a × ar = 64 or, a3 = 43 ⇒ a = 4. Again, a r + ar = 10 or, 4 r + 4r = 10 or, 4 + 4r2 = 10r or, 4r2 – 10r + 4 = 0 or, 2(2r2 – 5r + 2) = 0 or, 2r2 – 4r – r + 2 = 0 or, 2r(r – 2) – 1(r – 2) = 0 or, (r – 2) (2r – 1) = 0 Either, r – 2 = 0 ⇒ r = 2 Or, 2r – 1 = 0 ⇒ r = 1 2 Hence, the required three terms are 4 2, 4, 4 × 2 = 2, 4, 8. or, 4 1/2 , 4, 4 × 1 2 = 8, 4, 2. Assumption of Terms in GP SN Property Terms in GP 1. Three consecutive terms in GP a r , a, ar 2. Five successive terms in GP a r 2, a r , a, ar, ar2 3. Four consecutive terms in GP a r 3, a r , ar, ar3 4. Six consecutive terms in GP a r 5, a r 3, a r , ar, ar3 , ar5 We can solve the quadratic equations by factorization, perfect squaring, using formula and graphical methods. 228 Allied The Leading Mathematics-9 Sequence and Series 229

Algebra Algebra PRACTICE 8.4 Keeping Skill Sharp 1. (a) Write the definition of geometric series. (b) What is the general term of a sequence ? (c) Write any one property of the geometric sequence. 2. Which of the following sequences is a geometric sequence or series? (a) 1, 3, 9, 27, …….. (b) 2, 4, 6, 8, ……. Check Your Performance 3. (i) The first term and the common ratio of a geometric sequence are 32 and 1 2 respectively. (a) How can you find the common ratio of a geometric sequence? (b) Find the first six terms of the sequence. (ii) If the 6th term of a GP with the common ratio 3 is 243 2 , find its first term. 4. Observe the following geometric sequences and answer the following questions. (i) 243, 81, 27,…… (ii) 1 243 + 1 81 + 1 27 + ……… (a) Write the formula to find the general term of a geometric sequence. (b) Find the common ratio of the sequence. (c) Find the nth term of the sequence. (d) Find the 15th term of the sequence. 5. (i) The last term of a geometric sequence is 26244. If its first term and common ratio are 4 and 3 respectively then find the number of terms. (ii) Does 0.50625 belong to the geometric series 0.1, 0.15, 0.225……….. ? 6. (i) (x + 2), (x + 10) and 9(x + 2) are three consecutive terms in GP. (ii) (x + 1), (x – 5) and 1 are in a geometric series. (a) Find the value of x. (b) Find these three terms. (c) Compute the next three terms. (d) Find its nth term. 7. (i) The 6 times of the fifth term of a GP is 7500. If its common ratio is 5 then find its first term. (ii) The 4 times of the fourth term is equal to the one-fourth of the sixth term. If the first term is 1 512, find its common ratio. 8. (i) The third term and the seventh term of a GS are 4 and 64 respectively. (a) Find the first term of the GS. (b) Find common ratio of the GS. 230 Allied The Leading Mathematics-9 Sequence and Series 231

Algebra Algebra (ii) The fifth term and the ninth term of a GP are 8 and 2048 respectively. (a) Find the first term of the GS. (b) Find the common ratio of the GS. (c) Find the eleventh termof the GS. 9. (i) If the nth terms of the geometric sequences 1 32, 1 8, 1 2, .... and 4, 8, 16,… are equal, find the number of terms. (ii) The nth terms of the geometric sequences 1, 3, 9, ... and 1 6561, 1 729, 1 81, ... are equal. Find the value of n. 10. (i) Three successive numbers are in a GP and its product is 512. Their sum is 42. (ii) The sum of three consecutive terms in GP is 62 and their product is 1000. (a) Suppose the three consecutive numbers in AP. (b) Find these three numbers. 11. (i) In a geometric sequence, the fifth term is 8 times of the second term and the fourth term is 24. (a) Find the first term of the GS. (b) Find common ratio of the GS. (c) Find the geometric series. (ii) Three numbers are in the ratio of 1:4:13. If 1 is added to all three numbers then they form a GP. Find the original numbers. 12. (i) A rich man saves Re. 1 in the first day, Rs. 2 in the second day, Rs. 4 in the third day, Rs. 8 in the fourth day and so on. How much money does he save in the 30th day? (ii) A Ping-Pong ball is dropped from a height of 20 meter on to a horizontal roof of a house and always reduces 1 8 of the distance of the previous fall in each bounce. What height does it fall at 7th time? 3. (i) (b) 32, 16, 8, 4, 2, 1 (ii) 1 2 4. (i) (b) 1 3 (c) 1 3 n – 6 (d) 1 19683 (ii) (b) 3 (c) 3n-6 (d) 19683 5. (i) 9 (ii) Yes, 5 6. (i) (a) If a = 2, (b) 4, 12, 36 (c) 108, 324, 972 (d) 4(3)n – 1 If a = – 4, (b) –2, –6, –18 (c) –54, –162, –486 (d) –2(3)n – 1 (ii) (a) If a = 3, (b) 4, –2, 1 (c) – 1 2, 1 4, – 1 8 (d) 4 – 1 2 n – 1 If a = 8, (b) 9, 3, 1 (c) 1 3, 1 9, 1 27 (d) 9 1 3 n – 1 7. (i) 2 (ii) 4 8. (i) (a) 1 (b) 2 (ii) (a) 1 32 (b) 4 (c) 22n – 7 (d) 32768 9. (i) 8 (ii) 9 10. (i) 2, 8, 32, or 32, 8, 2 (ii) 2, 10, 50 or 50, 10, 2 11. (i) (a) 3 (b) 2 (c) 3 + 6 + 12 + 24 + ... (ii) 2, 8, 26 12. (i) Rs. 536870912 (ii) 8.98 m Answers 230 Allied The Leading Mathematics-9 Sequence and Series 231

Algebra Algebra 9.1 Factorization in the Form of (a ± b)3 and a3 ± b3 At the end of this topic, the students will be able to: ¾ factorise the algebraic expressions in the form of (a ± b)3 and a3 ± b3 . Learning Objectives Review on Algebraic Expressions Observe the following figures and discuss. (i) (ii) (iii) (iv) (v) (vi) I J A a2 b2 ab ab a + b a + b a a a a b b b b F E D B G C I J A b2b(a – b) a – b a – b a – b (a – b)2 a – b (a – b)b b b b b F E D B G C a a a + b I J A a2 a – b a – b b2 ab b b b b ab F E D B G C a a a CHAPTER 9 FACTORIZATION 232 Allied The Leading Mathematics-9 Algebra 233

Algebra Algebra In the figure (i), Area of ABCD = (a + b)2 = a2 + ab + ab + b2 = a2 + 2ab + b2 i.e., a2 + 2ab + b2 = (a + b)2 => a2 + b2 = (a + b)2 – 2ab In the figure (ii), Area of AEFI = (a – b)2 = a2 − (a − b)b − b(a − b) – b2 = a2 − 2ab + b2 i.e., a2 − 2ab + b2 = (a − b)2 => a2 + b2 = (a − b)2 + 2ab Furthermore, (a + b)2 = (a − b)2 + 4ab and (a − b)2 = (a + b)2 − 4ab In the figure (iii), Area of AEFI = (a + b)(a − b) = a(a + b) − (a + b)b = a2 + ab − ab − b2 = a2 − b2 i.e., a2 − b2 = (a + b)(a − b). In the figure (iv), Area of shaded portion = a2 − b2 = (a + b)(a − b) i.e., a2 − b2 = (a + b)(a − b). In the figure (v), Area of rectangle = (x + 3)(x + 2) = x2 + 5x + 6 i.e., x2 + 5x + 6 = (x + 3)(x + 2). In the figure (vi), Area of rectangle = (x − 4)(x − 1) = x2 − 5x + 4 i.e., x2 − 5x + 4 = (x − 4)(x − 1). Geometric Interpretation of (a + b)3 A cube of length (a + b) is cut down in the length b on length, breadth and height as shown in the figure below and we get 8 pieces in which 2 pieces of cubes of the length a and b, 3/3 pieces of cuboids of a2 b and ab2 . a a a + b a + b a + b a b b b a a a a a b b b b a a2 b a2 b ab2 b3 a3 ab2 ab2 a2 b b b a2 b a2 b a2 b ab2 ab2 ab2 b3 a3 a3 + + + 3a2 b 3ab2 b3 (a + b)3 = a3 + 3a2 b + 3ab2 + b3 Hence, the expression (a + b)3 is the volume of a cube of length a + b. i.e., Vol. of cube = (a + b)(a + b)(a + b) = (a + b)3 232 Allied The Leading Mathematics-9 Algebra 233

Algebra Algebra Combining all there we have, (a + b)3 = a3 + 3a2 b + 3ab2 + b3 or, (a + b)3 = a3 + b3 + 3ab (a + b) ‘Alternatively’ We have, by distributive property (a + b)3 = (a + b) (a + b)2 = (a + b) (a2 + 2ab + b2 ) = a(a2 + 2ab + b2 ) + b(a2 + 2ab + b2 ) = a3 + 2a2 b + ab2 + ba2 + 2ab2 + b3 ∴ (a + b)3 = a3 + 3a2 b + 3ab2 + b3 or, (a + b)3 = a3 + b3 + 3ab(a + b) ∴ a3 + b3 = (a3 + b)3 – 3ab(a + b) Furthermore, for the factors of a3 + b3 ; a3 + b3 = a3 + 3a2 b + 3ab2 + b3 – 3a2 b – 3ab2 = (a + b)3 – 3ab(a + b) = (a + b) [(a + b)2 – 3ab] = (a + b) (a2 + 2ab + b2 – 3ab) = (a + b) (a2 – ab + b2 ) ∴ a3 + b3 = (a + b)(a2 – ab + b2 ) Hence, the factors a3 + b3 of are (a + b) and (a2 – ab + b2 ). Geometric Interpretation of (a – b)3 A cube of length a is cut down in the length b on length, breadth and height as shown in the figure below and we get 8 pieces in which 2 pieces of cubes of the length (a – b) and b, 3 pieces of cuboids of (a – b) 2 b and 3 pieces of cuboids of (a – b)b2 . a – ba – b a – b b b b a – b b a – b a – b b a – b a – b b a – b b b a – b b b a – b b b a – b a a a a a a – b a – b a – b b b b b Then, from the figure, a3 = (a – b) 3 + b3 + 3(a – b) 2 b + 3(a – b)b2 = (a – b)3 + b3 + 3b(a2 – 2ab + b2 ) + 3ab2 – 3b3 = (a – b)3 + b3 + 3a2 b – 6ab2 + 3b3 + 3ab2 – 3b3 = (a – b)3 + b3 + 3a2 b – 3ab2 ∴ (a – b)3 = a3 – 3a2 b + 3ab2 – b3 234 Allied The Leading Mathematics-9 Algebra 235

Algebra Algebra ‘Alternatively’ (a – b)3 = (a – b) (a – b)2 = (a – b) (a2 – 2ab + b2 ) = a(a2 – 2ab + b2 ) – b(a2 – 2ab + b2 ) = a3 – 2a2 b + ab2 – a2 b + 2ab2 – b3 \ (a – b)3 = a3 – 3a2 b + 3ab2 – b3 or, (a – b)3 = a3 – b3 – 3ab(a – b) ∴ a3 – b3 = (a – b)3 + 3ab(a – b) Furthermore, for the factors of a3 + b3 ; a3 – b3 = a3 – 3a2 b + 3ab2 – b3 + 3a2 b – 3ab2 = (a – b)3 + 3ab (a – b) = (a – b) [(a – b)2 + 3ab] = (a – b) (a2 – 2ab + b2 + 3ab) = (a – b) (a2 + ab + b2 ) ‘Alternatively’ a3 – b3 = (a – b)3 + 3ab(a – b) = (a – b){(a – b)2 + 3ab} = (a – b){a2 – 2ab + b2 + 3ab} = (a – b) (a2 + ab + b2 ) ∴ a3 – b3 = (a – b)(a2 + ab + b2 ) Hence, the factors a3 – b3 of are (a – b) and (a2 + ab + b2 ). Example-1 Find the cube of (3x + 2y) by using formula. Solution: Here, (3x + 2y)3 = (3x)3 + 3(3x)2 (2y) + 3(3x) (2y)2 + (2y)3 = 27x3 + 54x2 y + 36xy2 + 8y3 Example-2 Expand: (2x – 3y)3 Solution: Here, (2x – 3y)3 = (2x)3 – 3(2x)2 (3y) + 3(2x) (3y)2 – (3y)3 = 8x3 – 36x2 y + 54xy2 – 27y3 Example-3 Factorize: (a) (a – 2b)3 (b) 64m3 – 48m2 n + 12mn2 – n3 Solution: Here, (a) (a – 2b)3 = (a – 2b) (a – 2b) (a – 2b) (b) 64m3 – 48m2 n + 12mn2 – n3 = (4m)3 – 3(4m)2 (n) + 3(4m)(n)2 – (n)3 = (4m – n)3 = (4m – n)(4m – n) (4m – n). 234 Allied The Leading Mathematics-9 Algebra 235

Algebra Algebra Example-4 Factorize: (a) x3 + 8y3 (b) 27a3 b3 – 64c3 Solution: Here, (a) x3 + 8y3 = (x)3 + (2y)3 = (x + 2y) {(x)2 – x.2y + (2y)2 } [ a3 + b3 = (a + b)(a2 – ab + b2 )] = (x + 2y) (x2 – 2xy + 4y2 ) (b) 27a3 – 64b3 = (3a)3 – (4b)3 = (3a – 4b) {(3a)2 – 3a.4b + (4b)2 } [ a3 – b3 = (a – b)(a2 + ab + b2 )] = (3a – 4b) (9a2 + 12ab + 16b2 ) PRACTICE 9.1 Read Think Understand Do Keeping Skill Sharp 1. (a) What is factorization of algebraic expression? (b) What are the factors of a2 – b2 ? (c) Write down the formula for (a + b)2 in expanded form. (d) What are the factors of a3 – b3 ? (e) What are the factors of a4 – b4 ? 2. Find the square of the following expressions by using formula. (a) (a + 3) (b) (2 + b) (c) (4 – x) (d) (2x – 3y) (e) (4m + 3) (f)   m – 1 2m  (g)   xy + 1 x   (h)   a2 – 1 3a2   (i)   p2 + 1 5pq2   3. Factorize by using formulae: (a) x2 – y2 (b) 9x2 – y2 (c) 4a2 – 9 (d)   m2 – 1 m2   (e)   a2 – 1 4a2   (f)   16p2 – 1 64a2  (g) p4 – q4 (h) 4a4 – 9b4 (i) 16x4 – 81y4 (j) 4x4 + y4 (k) 64a4 + 1 (l) 36a4 + 9 236 Allied The Leading Mathematics-9 Algebra 237

Algebra Algebra Check Your Performance 4. Find the cube of the following expressions by using formula. (a) (x + 2) (b) (3 + y) (c) (2 – y) (d) (3x – 2y) (e) (5m + 2n) (f)   2y – 1 2y  5. Expand: (a) (2x + y)3 (b) (x – 3y)3 (c) (2m – n)3 (d) (2p – 3q)3 (e)   x + 1 x  3 (f)   3m + 1 3m   3 6. Express as the square of binomial. (a) x2 + 2x + 1 (b) y2 + 2y + 1 (c) 4x2 – 4x + 1 (d) 4m2 – 16mn + 16n2 (e) 9x2 + 12xy + 4y2 (f) 16m2 – 72mn + 81n2 (g) y2 + 2 + 1 y (h) 4m2 – 2 + 1 4m2 7. Express as the cube of binomial. (a) a3 + 3a2 + 3a + 1 (b) 8x3 + 12a2 + 6a + 1 (c) 27m3 – 54m2 n + 36mn2 – 8n3 (d) 8p3 – 36p2 q + 54pq2 – 27q3 8. Factorize: (a) (x + 2y)2 (b) (3p + 2q)3 (c) 125a3 + 75a2 b + 15ab2 + b3 (d) 343x3 – 294x2 y + 84xy2 – 8y3 9. Simplify: (a) (a + 1) (a – 2)2 (b) (3x + 1) (2x – 1)2 (c) (2a + 5) (a – 4)2 (d) –6x (2x + 9) (2x – 9) (e) (2y + 5) (3y + 2) (2y – 5) (f) (3x + 2y) (3x – 2y) (9x2 – 4y2 ) 10. Find the volume of the following cuboids. (a) x + 2y (b) 2x – y (c) 3x – 1 236 Allied The Leading Mathematics-9 Algebra 237

Algebra Algebra 11. Find the square root of : (a) 4x2 + 4x + 1 (b) 9p2 + 24pq + 16q (c) 4x2 – 2 + 1 4x2 (d) 1 a2 – 1 ab + 1 4b2 (e) 16 25a2 + 2ab + 25 16 b2 (f) y2 + 2 + 1 y2 4. (a) x3 + 6x2 + 12x + 8 (b) 27 + 27y + 9y2 + y3 (c) 8 – 12y + 6y2 – y3 (d) 27x3 – 54x2 y + 36xy = 8y3 (e) 125m3 + 150m2n + 60mn2 + 8n3 (f) 8y3 – 6y + 3 2y – 1 8y3 5. (a) 8x3 + 12x2 y + 6xy2 + y3 (b) x3 – 9x2 y + 27xy2 – 27y3 (c) 8m3 – 12m2 n + 6mn2 – n3 (d) 8p2 – 36p2 q + 54pq2 – 27q2 (e) x3 + 3x + 1 x + 1 x3 (f) 27m3 + 9m + 1 m + 1 27m3 6. (a) (x + 1)2 (b) (y + 1)2 (c) (2x – 1)2 (d) (2m – 4n)2 (e) (3x + 2y)2 (f) (4m – 9n)2 (g) (y + 1 y) 2 (h) (2m – 1 2m) 2 7. (a) (a + 1)3 (b) (2x + 1)3 (c) (3m – 2n)3 (d) (2p – 3q)3 8. (a) (x + 2y) (x + 2y) (b) (3p + 2q)(3p + 2q) (3p + 2q) (c) (5a + b) (5a + b) (5a + b) (d) (7x – 2y)(7x – 2y) (7x – 2y) 9. (a) a3 – 3a2 + 4 (b) 12x3 – 8x2 – x + 1 (c) 2a3 – 11a2 – 8a + 80 (d) 486x – 24x3 (e) 12y3 + 8y2 – 75y – 50 (f) 81x4 – 72x2 y2 + 16y4 10. (a) x3 + 6x2 y + 12xy2 + 8y3 (b) 8x2 – 12xy + 6xy2 – y3 (c) 27x3 – 27x2 + 9x – 1 11. (a) (2x +1) (b) (3p + 4q) (c) 2x – 1 2x (d) 1 a – 1 2b (e) 4a 5 + 5b 4 (f) y + 1 y Answers 238 Allied The Leading Mathematics-9 Algebra 239

Algebra Algebra 9.2 Factorization in Form of (a4 + a2b2 + b4) At the end of this topic, the students will be able to: ¾ factorise the algebraic expressions in the form of (a4 + a2 b2 + b4 ). Learning Objectives Factorization of Algebraic Expressions in the form of a4 + 4b4 and a4 + a2 b2 + b4 i) a4 + a2 b2 + b4 = (a2 ) 2 + 2a2 b2 + (b2 ) 2 – a2 b2 = (a2 + b2 ) 2 – (ab) 2 [ a2 + 2ab + b2 = (a + b) 2 ] = (a2 + b2 + ab) (a2 + b2 – ab) [ a2 – b2 = (a + b)(a – b)] = (a2 + ab + b2 ) (a2 – ab + b2 ) "Alternatively" a4 + a2 b2 + b4 = a4 + b4 + a2 b2 = (a2 ) 2 + (b2 ) 2 + a2 b2 = (a2 + b2 ) 2 – 2.a2 .b2 + a2 b2 [ a2 + b2 = (a + b) 2 – 2ab)] = (a2 + b2 ) 2 – (ab) 2 = (a2 + b2 + ab) (a2 + b2 – ab) [ a2 – b2 = (a + b)(a – b)] = (a2 + ab + b2 ) (a2 – ab + b2 ) ∴ a4 + a2 b2 + b4 = (a2 + ab + b2 ) (a2 – ab + b2 ). ii) a4 + 4b4 = (a2 ) 2 + 2a2 .2b2 + (2b2 ) 2 – 4a2 b2 = (a2 + 2b2 ) 2 – (2ab) 2 [ a2 + 2ab + b2 = (a + b) 2 ] = (a2 + 2b2 + 2ab) (a2 + 2b2 – 2ab) [ a2 – b2 = (a + b)(a – b)] = (a2 + 2ab + 2b2 ) (a2 – 2ab + 2b2 ) ∴ a4 + 4b4 = (a2 + 2ab + 2b2 ) (a2 – 2ab + 2b2 ). Example-1 Factorize : 1 + 64a4 Solution: Here, 1 + 64a4 = 12 + (8a2 ) 2 = (1 + 8a2 ) 2 – 2.1.8a2 [ a2 + b2 = (a + b)2 – 2ab] = (1 + 8a2 ) 2 – 16a2 = (1 + 8a2 ) 2 – (4a)2 = (1 + 8a2 + 4a) (1 + 8a2 – 4a) [ a2 – b2 = (a + b) (a – b)] 238 Allied The Leading Mathematics-9 Algebra 239

Algebra Algebra = (1 + 4a + 8a2 ) (1 – 4a + 8a2 ) ∴ 1 + 64a4 = (1 + 4a + 8a2 ) (1 – 4a + 8a) Example-2 Factorize M x4 + x2 + 1 Solution : Here, x4 + x2 + 1 = (x2 ) 2 + 12 + x2 = (x2 + 1)2 – 2x2 . 1 + x2 [ a2 + b2 = (a + b) 2 – 2ab)] = (x2 + 1)2 – x2 = (x2 + 1 + x) ( x2 + 1 – x) [ a2 – b2 = (a + b)(a –b)] = (x2 + x + 1) ( x2 – x + 1) ∴ x4 + x2 + 1 = (x2 + x + 1) ( x2 – x + 1) Example-3 Resolve into factors: a4 + 5a2 + 9 Solution : Here, a4 + 5a2 + 9 = (a2 ) 2 + (3)2 + 5a2 = (a2 + 3)2 – 2.a2 .3 + 5a2 [ a2 + b2 = (a + b) 2 – 2ab)] = (a2 + 3)2 – 6a2 + 5a2 = (a2 + 3)2 – a2 = (a2 + 3 + a) (a2 + 3 – a) [ a2 – b2 = (a + b)(a – b)] = (a2 + a + 3) (a2 – a + 3). Example-4 Resolve into factors: x4 + 3x2 – 4 Solution: Here, x4 + 3x2 – 4 = (x2 ) 2 + 4x2 – x2 – 4 = x2 (x2 + 4) – 1(x2 + 4) = (x2 – 1) (x2 + 4) = (x2 – 12 ) (x2 + 4) = (x + 1) (x – 1) (x2 + 4) [ a2 – b2 = (a + b)(a – b)] Example-5 Factorize : a2 + b2 – c2 – d2 – 2ab – 2cd Solution: Here, a2 + b2 – c2 – d2 – 2ab – 2cd = a2 – 2ab + b2 – c2 – 2cd – d2 = (a – b)2 – [c2 + 2cd + d2 ] = (a – b)2 – (c + d)2 = [(a – b) + (c + d)] [(a – b) – (c + d)] = (a – b + c + d) (a – b – c – d) Example-6 Factorize : 4a4 – 8a2 b2 + b4 Solution: Here, 4a4 – 8a2 b2 + b4 = (2a2 ) 2 + (b2 ) 2 – 8a2 b2 240 Allied The Leading Mathematics-9 Algebra 241

Algebra Algebra = (2a2 – b2 ) 2 + 2.2a2 .b2 – 8a2 b2 [ a2 + b2 = (a + b) 2 – 2ab)] = (2a2 – b2 ) 2 + 4a2 .b2 – 8a2 b2 = (2a2 – b2 ) 2 – (2ab)2 = (2a2 + 2ab – b2 ) (2a2 – 2ab – b2 ) [ a2 – b2 = (a + b) (a – b)] Example-7 Factorize: (a + b) 2 – 4(a + b) + 3 – b2 + 2b Solution : Here, (a + b)2 – 4(a + b) + 3 – b2 + 2b = (a + b)2 – 2(a + b).2 + 4 – b2 + 2b – 1 = (a + b – 2)2 – (b – 1)2 [ a2 – 2ab + b2 = (a – b)2 ] = (a + b – 2 + b – 1) (a + b – 2 – b + 1) [ a2 – b2 = (a + b)(a – b)] = (a + 2b – 3) (a – 1). Example-8 Factorize M 4a5 + 35a3 b2 + 121ab4 Solution : Here, 4a5 + 35a3 b2 + 121ab4 = a(4a4 + 35a2 b2 + 121b4 ) = a{(2a2 ) 2 + (11b2 ) 2 + 35a2 b2 } = a{(2a2 + 11b2 ) 2 – 2 . 2a2 . 11b2 + 35 a2 b2 } [ a2 + b2 = (a + b)2 – 2ab] = a{(2a2 + 11b2 ) 2 – 44a2 b2 + 35a2 b2 ] = a{(2a2 + 11b2 ) 2 – 9a2 b2 } = a{(2a2 + 11b2 ) 2 – (3ab)2 } = a(2a2 + 11b2 + 3ab)(2a2 + 11b2 – 3ab) [a2 – b2 = (a + b) (a – b)] Example-9 Find the factors of x4 y4 + 1 + y4 x4 Solution: Here, x4 y4 + y4 x4 + 1 = x2 y2 2 + y2 x2 2 + 1 = x2 y2 + y2 x2 2 – 2x2 y2 . y2 x2 + 1 = x2 y2 + y2 x2 2 – 12 = x2 y2 + 1 + y2 x2 x2 y2 – 1 + y2 x2 Example-10 Factorize : a4 b4 + 1 – 7a2 b2 Solution: Here, a4 b4 + 1 – 7a2 b2 = a2 b2 2 + 12 – 7a2 b2 240 Allied The Leading Mathematics-9 Algebra 241

Algebra Algebra = a2 b2 + 1 2 – 9a2 b2 = a2 b2 + 1 2 – 3a b 2 = a2 b2 + 1 + 3a b a2 b2 + 1 – 3a b Example-11 Factorize: a4 + 4b4 – 9c4 – 16d4 + 4a2 b2 – 24c2 d2 Solution: Here, a4 + 4b4 – 9c4 – 16d4 + 4a2 b2 – 24c2 d2 = a4 + 4a2 b2 + 4b4 – 9c4 – 24c2 d2 – 16d4 = (a2 ) 2 + 2.a2 .2b2 + (2b2 ) 2 – [(3c2 ) 2 + 2.3c2 .4d2 + (4d2 ) 2 ] = (a2 + 2b2 ) – (3c2 + 4d2 ) 2 = (a2 + 2b2 + 3c2 + 4d2 ) [a2 + 2b2 – (3c2 + 4d2 )] = (a2 + 2b2 + 3c2 + 4d2 ) (a2 + 2b2 – 3c2 – 4d2 ) PRACTICE 9.2 Read Think Understand Do Keeping Skill Sharp 1. (a) Write down the factors of x4 + 4y4 . (b) What are the factors of x4 +x2 y2 + y4 ? Check Your Performance 2. Factorize: (a) a4 – 1 (b) a4 – 16 (c) x6 – y4 (d) (x + y) 4 – (x – y) 4 (e) 4y5 + 81y (f) (x + y) 4 + 4(x – y) 4 3. Resolve into factors: (a) 1 + b2 + b4 (b) x4 + 3x2 + 4 (c) 4x4 + 3x2 + 1 (d) a4 – 31a2 + 9 (e) x8 + x4 y4 + y8 (f) x8 + 9x4 + 81 4. Factorize: (a) a4 + 5a2 + 4 (b) a4 – 5a2 + 4 (c) x4 + 7x2 – 18 (d) a4 – 5a2 – 6 (e) 3x4 + 7x2 + 4 (f) 4a4 – 5a2 + 1 5. Resolve into factors: (a) x2 + 2xy + y2 – a2 + 2ab – b2 (b) x2 + 4xy + 4y2 – 1 + 2b – b2 (c) a2 – 4a + 3 – b2 + 2b (d) x2 – 10x + 24 + 6y – 9y2 (e) (a + b) 4 + 4(a + b) 2 (a – b) 2 + 4(a – b) 4 – 4(a4 + 4a2 b2 + 4b4 ) 242 Allied The Leading Mathematics-9 Algebra 243

Algebra Algebra 6. Factorize: (a) a6 + a4 + a2 (b) 2a4 b + 2a2 b3 + 2b5 (c) x5 y – 39x3 y + 25xy (d) 2x6 – 86x4 + 18x2 (e) 98a5 – 92a3 b2 + 18ab4 (f) 12a5 b + 105a3 b + 363ab (g) x2 y2 + 1 + y2 x2 (h) x4 + 1 + 1 x4 (i) a4 b4 + 1 + a2 b2 (j) 4p4 q4 + q4 p4 (k) a4 b4 – 3 + 9b4 a4 (l) x4 – 57x2 + 256 7. Factorize: (a) 25p4 + 36q4 – 49r4 – 64s4 + 60p2 q2 – 112r2 s2 (b) a4 – a3 + a2 b2 – 2a2 b – 2ab2 – b3 + b4 2. (a) (a – 1) (a + 1) (a2 + 1) (b) (a – 2) (a + 2) (a2 + 4) (c) (x3 – y2 ) (x3 + y2 ) (d) 8xy (x2 + y2 ) (e) y(2y2 + 6y + 9) (2y2 – 6y + 9) (f) (5x2 – 2xy + y2 ) (x2 – 2xy + 5y2 ) 3. (a) (b2 + b + 1) (b2 – b + 1) (b) (x2 + x + 2) (x2 – x + 2) (c) (2x2 + x + 1) (2x2 – x + 1) (d) (a2 + 5a – 3) (a2 – 5a – 3) (e) (x4 – x2 y2 + y4 ) (x2 + xy + y2 ) (x2 – xy + y2 ) (f) (x4 + 3x2 + 9) (x2 + 3x + 3) (x2 – 3x + 3) 4. (a) (a2 + 4) (a2 + 1) (b) (a – 2) (a + 2) (a – 1) (a + 1) (c) (x2 + 9) (x2 – 2) (d) (a2 – 6) (a2 + 1) (e) (3x2 + 4) (x2 + 1) (f) (a2 – 1) (4a2 – 1) 5. (a) (x + y – a + b) (x + y + a – b) (b) (x + 2x + b – 1) (x + 2y – b + 1) (c) (a + b – 3) (a – b –1) (d) (x + 3y – 6) (x – 3y – 4) (e) (a2 – 2ab – b2 ) (5a2 – 2ab + 7b2 ) 6. (a) a2 (a2 + a + 1) (a2 – a + 1) (b) 2b (a2 + ab + b2 ) (a2 – ab + b2 ) (c) xy(x2 + 5 + 7y) (x2 + 5 – 7y) (d) 2x2 (x2 – 7x + 3), (x2 + 7x + 3) (e) 2a(7a2 – 2ab – 3b) (7a2 + 2ab – 3b) (f) 3ab(2a2 – 3a + 11) (2a2 + 3a + 11) (g) x y + 1 + y x x y – 1 + y x (h) x2 + 1 + 1 x2 x2 – 1 + 1 x2 (i) a2 b2 + a b + 1 a2 b2 – a b + 1 (j) 2p2 q2 + 2 + q p 2p2 q2 – 2 + q p (k) a2 b2 + 3 + 3b2 a2 a2 b2 – 3 + 3b2 a2 (l) (x2 + 5x – 16) (x2 – 5x – 16) 7. (a) (5p2 + 6q2 – 7r2 – 8s2 ) (5p2 + 6q2 + 7r2 + 8s2 ) (b) (a2 + ab + b2 ) (a2 – ab + b2 – a – b) Answers Project Work Derive the formulae of (a + b)2 and a2 – b2 by cutting the thick paper. 242 Allied The Leading Mathematics-9 Algebra 243

Algebra Algebra 10.1 Highest Common Factor (HCF) At the end of this topic, the students will be able to: ¾ find the HCF from the given algebraic expressions. Learning Objectives Consider the polynomials, x3 – 3x2 and x3 – 6x2 + 9x Now, factorizing the given polynomials; x3 – 3x2 = x2 (x – 3) x3 – 6x2 + 9x = x (x2 – 6x + 9) = x(x – 3)2 Observe the factors of each polynomial and think if you can find polynomials that divide both of the given polynomials. The polynomials that divide the given polynomials are x, (x – 3) and x(x – 3). (why?) The polynomials x, (x – 3), and x(x – 3) are the common factors of the given polynomials. Among these three polynomials which one has the highest degree? x(x – 3) has hight degree, which is the HCF of the given polynomials x2 – 3x and x3 – 6x2 + 9x. HCF of polynomials is a polynomial of the highest degree of the common factors of the given polynomials, which exactly divides the given polynomials. So, it is also called greatest common divisor (GCD or gcd). Method of Finding HCF At first, the given expressions are polynomials with integral coefficients, we first resolve each of the given expressions into factors, determine the factors common to each of them and then finally multiply all the common factors to get the required HCF. ∴ HCF = Common factor/s of the given expressionsin the form of factors. Note: If there is no common factor, then consider 1 is the HCF. (Why ?) For this, we need the following algebraic formulae and relations to arrange the given expressions into factors: (i) a2 ± 2ab + b2 = (a ± b)2 x x(x – 3) Shaded portion = Product of Common Factors = HCF (x – 3) CHAPTER 10 HCF AND LCM 244 Allied The Leading Mathematics-9 Algebra 245

Algebra Algebra (ii) a2 – b2 = (a + b) (a – b) (iii) a2 + b2 = (a ± b)2 2ab, should be converted into the form a2 – b2 . (iv) a3 ± b3 = (a ± b) (a2 ab + b2 ) (v) a4 – b4 = (a2 + b2 ) (a + b)(a – b) (vi) a4 + ab + b4 = (a2 + ab + b2 ) (a2 – ab + b2 ) (vii) a6 – b6 = (a + b)(a – b) (a2 + ab + b2 ) (a2 – ab + b2 ) (viii) ax2 + bx + c = ax2 + px + qx + c, where b = p + q and a × c = p × q, and taking common. (ix) (a ± b)3 = a3 ± 3a2 b + 3ab2 ± b3 = a3 ± b3 ± 3ab(a ± b) Example-1 Find the HCF of the following: x2 – 1, x3 + 1 and x3 – 1. Solution: Here, 1st expression = (x2 – 1) = (x + 1) (x – 1) 2nd expression = (x3 + 1) = (x + 1) (x2 – x + 1) 3rd expression = (x3 – 1) = (x – 1) (x2 + x + 1) Since there is non-trivial common factor to all expression, so HCF = 1. Example-2 Find the HCF of the following: a3 – b3 and a3 + a2 b + ab2 . Solution: Here, 1st expression = a3 – b3 = (a – b)(a2 + ab + b2 ) 2nd expression = a3 + a2 b + ab2 = a(a2 + ab + b2 ) ∴ HCF = (a2 + ab + b2 ). Example-3 Find the HCF of: 6x2 – 18x – 60 and 3x2 + 12x + 12. Solution: Here, 1st expression = 6x2 – 18x – 60 = 6(x2 – 3x –10) = 2.3 (x – 5) (x + 2) 2nd expression = 3x2 + 12x + 12 = 3(x2 + 4x + 4) = 3 (x + 2)2 ∴ HCF = 3(x + 2). Example-4 Find the HCF of: 3m2 – 2mn – n2 , m2 – n2 , and m3 – n3 Solution: Here, 1st expression = 3m2 – 2mn – n2 = 3m2 – 3mn + mn – n2 = 3m(m – n) + n(m – n) = (m –n)(3m + n) 2nd expression = m2 – n2 = (m + n) (m – n) 3rd expression = m3 – n3 = (m – n) (m2 + mn + n2 ) ∴ HCF = (m – n) Example-5 Find HCF of x3 – 2x2 – 3x, x3 – 9x and x4 – 27x Solution: Here, 1st Exp. = x3 – 2x2 – 3x = x(x2 – 2x– 3) = x[x(x – 3) + 1 (x – 3)] = x(x – 3) (x +1) 244 Allied The Leading Mathematics-9 Algebra 245

Algebra Algebra 2nd Exp.= x3 – 9x = x(x2 – 9) = x(x + 3) (x – 3) 3rd Exp. = x4 – 27x = x(x3 – 33 ) = x(x – 3) (x2 + 3x + 9) ∴ HCF = x(x – 3) Example-6 Find HCF of : x3 + 1 x3 , x4 + 1 + 1 x4 and x3 + x + 1 x Solution: Here, 1st Exp. = x3 + 1 x3 = x + 1 x x2 – x . 1 x + 1 x2 = x + 1 x x2 – 1 + 1 x2 2nd Exp. = x4 + 1 + 1 x4 = (x2 ) 2 + 1 x2 2 + 1 = x2 + 1 x2 2 – 2x . 1 x2 + 1 = x2 + 1 x2 2 – 1 = x2 + 1 x2 + 1 x2 + 1 x2 – 1 = x2 + 1 + 1 x2 x2 – 1 + 1 x2 3rd Exp. = x3 + x + 1 x = x x2 + 1+ 1 x2 ∴ HCF = 1 Example-7 Find HCF of : x3 – 3x2 – x + 3 and x3 – x2 – 9x + 9 Solution: Here, 1st Exp. = x3 – 3x2 – x+3 = x2 (x – 3) – 1(x – 3) = (x – 3)(x2 – 1) = (x – 3)(x+1)(x – 1) 2nd Exp. = x3 – x2 – 9x + 9 = x2 (x – 1) – 9(x – 1) = (x – 1) (x2 – 9) = (x – 1) (x + 3) (x – 3) ∴ HCF = (x – 3)(x – 1) = (x2 – 4x + 3). Example-8 Find HCF of: x2 + y2 – z2 + 2xy, x2 – y2 + z2 + 2zx and – x2 + y2 + z2 + 2yz Solution: Here, 1st Exp. = x2 + y2 – z2 + 2xy = x2 + 2xy+ y2 – z2 = (x + y)2 – z2 = (x + y + z) (x + y – z) 2nd Exp. = x2 – y2 + z2 + 2zx = x2 + 2zx + z2 – y2 = (x + z)2 – y2 = (x + z + y) (x + z – y) = (x + y + z) (x – y + z) 246 Allied The Leading Mathematics-9 Algebra 247

Algebra Algebra 3rd Exp. = – x2 + y2 + z2 + 2yz = y2 + 2yz+ z2 – x2 = (y+ z)2 – x2 = (y+ z + x) (y+ z – x) = (x + y + z) (–x+ y + z) ∴ HCF = (x+ y + z). Example-9 Find HCF of 8a3 + 1 and16a4 – 4a2 + 4a – 1. Solution: Here, 1st Exp. = 8a3 + 1 = (2a) 3 + 13 = (2a +1) (4a2 – 2a +1) 2nd Exp. =16a4 – 4a2 + 4a – 1 = 16a4 – (4a2 – 4a +1) = 16a4 – [(2a) – 2.2a + 12 ] = (4a2 ) 2– (2a –1)= (4a2 + 2a –1) (4a2 – 2a +1) ∴ HCF = 4a2 – 2a +1 Example-10 Find HCF of: x4 + x2 + 1 and x3 + 2x2 + 2x + 1 Solution: Here, 1st Exp. = x4 + x2 + 1 = (x2 ) 2 + 12 + x2 = (x2 + 1)2 – x2 = (x2 + 1+ x) (x2 + 1– x) = (x2 + x +1) (x2 – x +1) 2nd Exp. = x3 + 2x2 + 2x + 1 = x3 +13 + 2x(x + 1) = (x + 1) (x2 – x +1) + 2x(x + 1) = (x + 1) (x2 – x +1+ 2x) = (x + 1) (x2 + x + 1) ∴ HCF = (x2 + x +1) PRACTICE 10.1 Read Think Understand Do Keeping Skill Sharp 1. (a) Define HCF of algebraic expressions. (b) If there is no common factor in the given algebraic expressions, what is the HCF? (c) Write the HCF of the three expressions having the same algebraic factors. (d) What is the HCF of (3x + y) (3x – 2) and (2x + y) (3x – 2)? Check Your Performance 2. Find the Highest Common Factor (HCF) of: (a) a2 + ab and ab + b2 (b) a2 – b2 and a2 – ab (c) a3 – 3a2 b and a2 – 9b2 (d) (x – 2)3 and x3 – 8 (e) 3ax(a – x)3 and 2a2 x(a – x)2 (f) 4x (x2 – y2 ) and 12x2 y (x2 + 3xy + 2y2 ) 246 Allied The Leading Mathematics-9 Algebra 247

Algebra Algebra 3. Find the Highest Common Factor (HCF) of: (a) x4 – x and x3 + x2 + x (b) a2 – 9 and a2 – 6a + 9 (c) a3 – b3 and a2 – b2 (d) 10x2 + 7x + 1 and 20x3 + 9x2 + x (e) x2 – 5x + 6 and x2 – 4 (f) a2 + 2ab + b2 – c2 and a2 – b2 + ac – bc 4. Find the Highest Common Factor (HCF) of: (a) 1 + 4x + 4x2 –16x4 and 1 + 2x–8x3 –16x4 (b) a4 + a2 b2 + b4 and a3 + 2a2 b + 2ab2 + b3 (c) a7 – 1 a5 and a4 + 1 + 1 a4 (d) x4 + x2 + 169 and x3 + x(x + 13) + 4x2 (e) x2 + (b – a)x – ab and x2 – (a – b)x – ab (f) x2 + (p – q)x – pq and x2 + x – q2 – q 5. Find the Highest Common Factor (HCF) of: (a) 2a4 – 32, a2 – 5a + 6 and a2 + 2a – 8 (b) a2 – a – 2, a2 + a – 6 and 3a2 – 13a + 14 (c) 3x3 + 6x2 – 9x, 9x3 – 9x, 5x3 – 5x (d) a4 + a2 b2 + b4 , a3 + b3 and a3 – a2 b + ab2 (e) x3 + 1 + 2x + x2 , x3 – 1 and x4 + x2 + 1 (f) 2x2 + 5x + 2, 3x2 + 8x + 4 and 2x2 + 3x – 2 2. (a) a + b (b) a – b (c) a – 3b (d) x – 2 (e) ax(a – x)2 (f) 4x(x + y) 3. (a) x(x2 + x + 1) (b) a – 3 (c) a – b (d) 5x + 1 (e) x – 2 (f) a + b + c 4. (a) 1 + 2x + 4x2 (b) a2 + ab + b2 (c) a2 + 1 + 1 a2 a2 – 1 + 1 a2 (d) x2 + 5x + 3 (e) (x – a) (x + b) (f) x – q 5. (a) a – 2 (b) a – 2 (c) x(x – 1) (d) a2 – ab + b2 (e) x2 + x + 1 (f) x + 2 Answers 248 Allied The Leading Mathematics-9 Algebra 249

Algebra Algebra 10.2 Lowest Common Multiple (LCM) At the end of this topic, the students will be able to: ¾ find the LCM of the given algebraic expressions. Learning Objectives Consider any two polynomials; (x2 – 4)2 and x2 – 3x + 2. Now, write down their factors as, (x2 – 4)2 = (x + 2)2 (x – 2)2 x2 – 3x + 2 = (x – 2) (x – 1) Can you form a polynomial that is divisible by the given polynomials? A polynomial that is divisible by the given polynomials should contain the factors of each of the given polynomials. So, the required polynomial divisible by the given polynomials is formed by the following factors: (x + 2)2 (x – 2)2 (x – 1) [ Here, you see the different bases of the two expressions with their highest power.] The product (x + 2)2 (x – 2)2 (x – 1) is divisible by the given expressions. (why?) Do you think there are many expressions divisible by the given expression? Obviously; one can form many products including these factors as, e.g. 3(x + 2)2 (x – 2)2 (x – 1); (x + 2)2 (x – 2)2 (x – 1) (x –3) (x + 2)2 (x – 2)2 (x – 1) (x2 – 2x + 3), etc. all are divisible by the given polynomials. (why?). Now, compare the degree and the constant factor of all the above discussed polynomials. (x + 2)2 (x – 2)2 (x – 1) ................................... degree 5 constant factor 1, 3(x + 2)2 (x – 2)2 (x – 1) ................................. degree 5 with constant factor 3, (x + 2)2 (x – 2)2 (x – 1) (x – 3) ........................ degree 6 with constant factor 1, (x + 2)2 (x – 2)2 (x – 1) (x2 – 2x + 3) .............. degree 7 with constant factor 1, etc. Among these products formed by the polynomials divisible by the given polynomials x2 – 4, and x2 – 3x + 2, the product (x + 2)2 (x – 2)2 (x – 1) has least degree and least constant factor. An algebraic expression of the lowest degree which is divided by each of the given expression is called lowest common multiple (LCM). Method of Finding LCM * First of all factorize the given expressions. * Take the common factors and then the remaining expressions which are not in common. * If there is no any common factor, then the LCM is the product of the given expressions. (x +2) (x +2) (x – 2) (x – 2) All shaded portions = Product of Common and Remaining Factors = LCM (x – 1) 248 Allied The Leading Mathematics-9 Algebra 249

Algebra Algebra LCM =Common factor ofthe given expressions ×Remaining factors = HCF ×Remaining factors. If there are only two expressions, HCF × LCM = 1st expression × 2nd expression. Example-1 Find the LCM of 2(a2 – b2 ) and 3a2 – 3ab. Solution: Here, 1st expression = 2(a2 – b2 ) = 2(a + b)(a – b) 2nd expression = 3a2 – 3ab = 3a(a – b) Hence, LCM = 2 × 3 × a × (a – b) (a + b) = 6a(a2 – b2 ). Example-2 Find the LCM of x2 – 3x + 2, x2 – 4x + 3 and x2 – 7x + 6. Solution: Here, 1st Exp. = x2 – 3x + 2 = x2 – 2x – x + 2 = x(x – 2) – 1(x – 2) = (x – 1)(x – 2) 2nd Exp. = x2 – 4x + 3 = x2 – 3x – x + 3 = x(x – 3) – 1(x – 3) = (x – 1)(x – 3) 3rd Exp. = x2 – 7x + 6 = x2 – 6x – x + 6 = x(x – 6) – 1(x – 6) = (x – 1)(x – 6) So, LCM = (x – 1)(x – 2)(x – 3) (x – 6). Example-3 Find the LCM: x3 – 64y3 , x2 – 6xy + 8y2 and x2 – 16y2 . Solution: Here, 1st expression = x3 – 64y3 = x3 – (4y)3 = (x – 4y) (x2 + 4xy + 16y2 ) 2nd expression = x2 – 6xy + 8y2 = x2 – 4xy – 2xy + 8y2 = x(x– 4y) – 2y(x – 4y) = (x – 4y) (x – 2y) 3rd expression = x2 – 16y2 = x2 – (4y)2 = (x + 4y) (x – 4y) ∴ LCM = (x – 4y) (x2 + 4xy + 16y2 ) (x – 2y) (x + 4y). Example-4 Find the LCM of 3a3 – 18a2 x + 27ax2 , 4a4 + 24a3 x + 36a2 x2 and 6a4 – 54a2 x2 . Solution: Here, 1st expression = 3a3 – 18a2 x + 27ax2 = 3a(a2 – 6ax + 9x2 ) = 3a(a – 3x)2 2nd expression = 4a4 + 24a3 x + 36a2 x2 = 4a2 (a2 + 6ax + 9x2 ) = 4a2 (a + 3x)2 3rd expression = 6a4 – 54a2 x2 = 6a2 (a2 – 9x2 ) = 6a2 (a + 3x) (a – 3x) Then, LCM = 3 × 2 × 2 × a2 (a + 3x)2 (a – 3x)2 = 12a2 (a + 3x)2 (a – 3x)2 . 250 Allied The Leading Mathematics-9 Algebra 251