Tamil Nadu State Board Class 10 Maths Textbook PDF

2 NUMBERS AND SEQUENCES “I know numbers are beautiful, if they aren’t beautiful, nothing is”...

2 NUMBERS AND SEQUENCES “I know numbers are beautiful, if they aren’t beautiful, nothing is” - Paul Erdos Srinivasa Ramanujan was an Indian mathematical genius who was born in Erode in a poor family. He was a child prodigy and made calculations at lightning speed. He produced thousands of precious formulae, jotting them on his three notebooks which are now preserved at the University of Madras. With the help of several notable men, he became the first research scholar in the mathematics department of University of Madras. Subsequently, Srinivasa Ramanujan he went to England and collaborated with G.H. Hardy for five (1887-1920) years from 1914 to 1919. He possessed great interest in observing the pattern of numbers and produced several new results in Analytic Number Theory. His mathematical ability was compared to Euler and Jacobi, the two great mathematicians of the past Era. Ramanujan wrote thirty important research papers and wrote seven research papers in collaboration with G.H. Hardy. He has produced 3972 formulas and theorems in very short span of 32 years lifetime. He was awarded B.A. degree for research in 1916 by Cambridge University which is equivalent to modern day Ph.D. Degree. For his contributions to number theory, he was made Fellow of Royal Society (F.R.S.) in 1918. His works continue to delight mathematicians worldwide even today. Many surprising connections are made in the last few years of work made by Ramanujan nearly a century ago. Learning Outcomes z To study the concept of Euclid’s Division Lemma. z To understand Euclid’s Division Algorithm. z To find the L.C.M and H.C.F using Euclid’s Division Algorithm. z To understand the Fundamental Theorem of Arithmetic. z To understand the congruence modulo ‘n’, addition modulo ‘n’ and multiplication modulo ‘n’. z To define sequence and to understand sequence as a function. z To define an Arithmetic Progression (A.P) and Geometric Progression (G.P). z To find the n th term of an A.P and its sum to n terms. z To find the n th term of a G.P and its sum to n terms. z To determine the sum of some finite series such as å n , å n 2 , å n 3. Numbers and Sequences 37 X Std_EM Final.indb 37 05-03-2019 17:58:07  2.1 Introduction The study of numbers has fascinated humans since several thousands of years. The discovery of Lebombo and Ishango bones which existed around 25000 years ago has confirmed the fact that humans made counting process for meeting various day to day needs. By making notches in the bones they carried out counting efficiently. Most consider that these bones were used as lunar calendar for knowing the phases of moon thereby understanding Number carvings in Ishango Bone the seasons. Thus the bones were considered to be Fig.2.1 the ancient tools for counting. We have come a long way since this primitive counting method existed. It is very true that the patterns exhibited by numbers have fascinated almost all professional mathematicians’ right from the time of Pythagoras to current time. We will be discussing significant concepts provided by Euclid and continue our journey of studying Modular Arithmetic and knowing about Sequences and Finite Series. These ideas are most fundamental to your progress in mathematics for upcoming classes. It is time for us to begin our journey to understand the most fascinating part of mathematics, namely, the study of numbers. 2.2 Euclid’s Division Lemma Euclid, one of the most important mathematicians wrote an important book named “Elements” in 13 volumes. The first six volumes were devoted to Geometry and for this reason, Euclid is called the “Father of Geometry”. But in the next few volumes, he made fundamental contributions to understand the properties of numbers. One among them is the “Euclid’s Divison Lemma”. This is a simplified version of the long division process that you were performing for division of numbers in earlier classes. Le us now discuss Euclid’s Lemma and its application through an Algorithm termed as “Euclid’s Division Algorithm”. Lemma is an auxiliary result used for proving an important theorem. It is usually considered as a mini theorem. Theorem 1: Euclid’s Division Lemma Let a and b (a > b ) be any two positive integers. Then, there exist unique integers q and r such that a = bq + r, 0 ≤ r < b. Note 1. The remainder is always less than the divisor. 2. If r = 0 then a = bq so b divides a. 3. Similarly, if b divides a then a = bq 38 10th Standard Mathematics X Std_EM Final.indb 38 05-03-2019 17:58:08  Example 2.1 We have 34 cakes. Each box can hold 5 cakes only. How many boxes we need to pack and how many cakes are unpacked? Solution We see that 6 boxes are required to pack 30 cakes with 4 cakes left over. This distribution of cakes can be understood as follows: 34 = 5 × 6 + 4 Total number = Number of cakes × Number of + Number of cakes of cakes in each box boxes left over Dividend = Divisor × Quotient + Remainder a b q r Note 1. The above lemma is nothing but a restatement of the long division process, the integers q and r are called quotient and remainder respectively. 2. When a positive integer is divided by 2 the remainder is either 0 or 1. So, any positive integer will of the form 2k, 2k+1 for some integer k. Euclid’s Division Lemma can be generalised to any two integers. Generalised form of Euclid’s division lemma. If a and b are any two integers then there exist unique integers q and r such that + r , where 0 ≤r b Step1: Using Euclid’s division lemma a = bq + r ; 0 ≤ r < b. where q is the quotient, r is the remainder. If r = 0 then b is the Highest Common Factor of a and b. Step 2: Otherwise applying Euclid’s division lemma divide b by r to get b = rq1 + r1, 0 ≤ r1 < r Step 3: If r1 = 0 then r is the Highest common factor of a and b. Step 4: Otherwise using Euclid’s division lemma, repeat the process until we get the remainder zero. In that case, the corresponding divisor is the H.C.F of a and b. Note 1. The above algorithm will always produce remainder zero at some stage. Hence the algorithm should terminate. 2. Euclid’s Division Algorithm is a repeated application of Division Lemma until we get zero remainder. 3. Highest Common Factor (H.C.F) of two positive numbers is denoted by (a,b). 4. Highest Common Factor (H.C.F.) is also called as Greatest Common Divisor (G.C.D.). 40 10th Standard Mathematics X Std_EM Final.indb 40 05-03-2019 17:58:13  Progress Check 1. Euclid’s division algorithm is a repeated application of division lemma until we get remainder as _____. 2. The H.C.F. of two equal positive integers k, k is _____. Illustration 1 Using the above Algorithm, let us find H.C.F. of two given positive integers. Let a = 273 and b = 119 be the two given positive integers such that a > b. We start dividing 273 by 119 using Euclid’s division lemma, we get 273 = 119 × 2 + 35 …(1) The remainder is 35 ¹ 0. Therefore, applying Euclid’s Division Algorithm to the divisor 119 and remainder 35, we get 119 = 35 × 3 + 14 …(2) The remainder is 14 ¹ 0. Applying Euclid’s Division Algorithm to the divisor 35 and remainder 14, we get 35 =14 ×2 +7 …(3) The remainder 7 is ≠ 0. Applying Euclid’s Division Algorithm to the divisor 14 and remainder 7. we get, 14 =7 × 2 +0 …(4) The remainder at this stage = 0. The divisor at this stage = 7. Therefore Highest Common Factor of 273, 119 = 7. Example 2.4 If the Highest Common Factor of 210 and 55 is expressible in the form 55x - 325 , find x. Solution Using Euclid’s Division Algorithm, let us find the HCF of given numbers 210 = 55 × 3 + 45 55 = 45 × 1 + 10 45 = 10 × 4 + 5 10 = 5 × 2 + 0 The remainder is zero. So, the last divisor 5 is the Highest Common Factor (H.C.F.) of 210 and 55. Since, H.C.F. is expressible in the form 55x − 325 = 5 gives 55x = 330 Hence x =6 Numbers and Sequences 41 X Std_EM Final.indb 41 05-03-2019 17:58:14  Example 2.5 Find the greatest number that will divide 445 and 572 leaving remainders 4 and 5 respectively. Solution Since the remainders are 4, 5 respectively the required number is the H.C.F. of the number 445 − 4 = 441, 572 − 5 = 567. Hence, we will determine the H.C.F of 441 and 567. Using Euclid’s Division Algorithm, we have 567 = 441 × 1 + 126 441 = 126 × 3 + 63 126 = 63 × 2 + 0 Therefore H.C.F. of 441, 567 = 63 and so the required number is 63. Activity 1 This activity helps you to find H.C.F. of two positive numbers. We first observe the following instructions. (i) Construct a rectangle whose length and breadth are the given numbers. (ii) Try to fill the rectangle using small squares. (iii) Try with 1×1 square; Try with 2×2 square; Try with 3 ´ 3 square and so on. (iv) The side of the largest square that can fill the whole rectangle without any gap will be H.C.F. of the given numbers. (v) Find the H.C.F of (a) 12,20 (b) 16,24 (c) 11,9 Theorem 3 If a, b are two positive integers with a > b then G.C.D of (a,b) = G.C.D of (a - b, b). Activity 2 This is another activity to determine H.C.F. of two given positive integers. (i) From the given numbers, subtract the smaller from the larger number. (ii) From the remaining numbers, subtract smaller from the larger. (iii) Repeat the subtraction process by subtracting smaller from the larger. (iv) Stop the process, when the numbers become equal. (v) The number representing equal numbers obtained in step (iv), will be the H.C.F. of the given numbers. Using this Activity, find the H.C.F. of (i) 90,15 (ii) 80,25 (iii) 40,16 (iv) 23,12 (v) 93,13 42 10th Standard Mathematics X Std_EM Final.indb 42 05-03-2019 17:58:15  Highest Common Factor of three numbers We can apply Euclid’s Division Algorithm twice to find the Highest Common Factor (H.C.F) of three positive integers using the following procedure. Let a, b, c be the given positive integers. (i) Find H.C.F of a,b. Call it as d d = (a,b) (ii) Find H.C.F of d and c. This will be the H.C.F of the three given numbers a, b, c Example 2.6 Find the H.C.F of 396, 504, 636. Solution To find H.C.F of three given numbers, first we have to find H.C.F of the first two numbers. To find H.C.F of 396 and 504 Using Euclid’s division algorithm we get 504 = 396 × 1 + 108 The remainder is 108 ¹ 0 Again applying Euclid’s division algorithm 396 = 108 × 3 + 72 The remainder is 72 ¹ 0, Again applying Euclid’s division algorithm 108 = 72 × 1 + 36 The remainder is 36 ¹ 0, Again applying Euclid’s division algorithm 72 = 36 × 2 + 0 Here the remainder is zero. Therefore H.C.F. of 396, 504 = 36. To find the H.C.F. of 636 and 36. Using Euclid’s division algorithm we get 636 = 36 × 17 + 24 The remainder is 24 ¹ 0 Again applying Euclid’s division algorithm 36 = 24 × 1 + 12 The remainder is 12 ¹ 0 Again applying Euclid’s division algorithm 24 = 12 × 2 + 0 Here the remainder is zero. Therefore H.C.F of 636,36 = 12 Therefore Highest Common Factor of 396, 504 and 636 is 12. Two positive integers are said to be relatively prime or co prime if their Highest Common Factor is 1. Exercise 2.1 1. Find all positive integers, when divided by 3 leaves remainder 2. 2. A man has 532 flower pots. He wants to arrange them in rows such that each row contains 21 flower pots. Find the number of completed rows and how many flower pots are left over. Numbers and Sequences 43 X Std_EM Final.indb 43 05-03-2019 17:58:17  3. Prove that the product of two consecutive positive integers is divisible by 2. 4. When the positive integers a , b and c are divided by 13, the respective remainders are 9,7 and 10. Show that a+b+c is divisible by 13. 5. Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4. 6. Use Euclid’s Division Algorithm to find the Highest Common Factor (H.C.F) of (i) 340 and 412 (ii) 867 and 255 (iii)10224 and 9648 (iv) 84, 90 and 120 7. Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case. 8. If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32 x + 60 y. 9. A positive integer when divided by 88 gives the remainder 61. What will be the remainder when the same number is divided by 11? 10. Prove that two consecutive positive integers are always coprime. 2.4 Fundamental Theorem of Arithmetic Let us consider the following conversation between a Teacher and students. Teacher : Factorise the number 240. Malar : 24 ×10 Raghu : 8×30 Iniya : 12×20 Kumar : 15×16 Malar : Whose answer is correct Sir? Teacher : All the answers are correct. Raghu : How sir? Teacher : Split each of the factors into product of prime numbers. Malar : 2×2×2×3×2×5 Raghu : 2×2×2×2×3×5 Iniya : 2×2×3×2×2×5 Kumar : 3×5×2×2×2×2 Teacher : Good! Now, count the number of 2’s, 3’s and 5’s. Malar : I got four 2’s, one 3 and one 5. Raghu : I got four 2’s, one 3 and one 5. Iniya : I also got the same numbers too. Kumar : Me too sir. Malar : All of us got four 2’s, one 3 and one 5. This is very surprising to us. Teacher : Yes, It should be. Once any number is factorized up to a product of prime numbers, everyone should get the same collection of prime numbers. This concept leads us to the following important theorem. 44 10th Standard Mathematics X Std_EM Final.indb 44 05-03-2019 17:58:18  Theorem 4 (Fundamental Theorem of Arithmetic) (without proof ) “Every natural number except 1 can be factorized as a product of primes and this factorization is unique except for the order in which the prime factors are written.” The fundamental theorem N asserts that every composite number can be decomposed as a product of prime numbers and that x1 x2 x3 the decomposition is unique. In the sense that there is one and only way to express the decomposition as y1 y2 y3 y4 y5 y6 product of primes................... In general, we conclude that q q q q given a composite number N, we p1 p2 p3 p4... q 1 2 3 4 pn n decompose it uniquely in the form Fig.2.2 N = p1 × p2 × p3 ×  × pn where p1, p2, p3,..., pn are primes and q1, q2, q 3,..., qn are natural q q1 q2 3 q n numbers. First, we try to factorize N into its factors. If all the Thinking Corner factors are themselves primes then we can stop. Otherwise, we try to further split the factors which are not prime. Is 1 a prime number? Continue the process till we get only prime numbers. Illustration Progress Check For example, if we try to factorize 32760 we get 1. Every natural number except ______ can be expressed as ______. 32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 2. In how many ways a composite = 23 × 32 × 51 × 71 × 131 number can be written as product of Thus, in whatever way we try to power of primes? factorize 32760, we should finally get three 3. The number of divisors of any prime 2’s, two 3’s, one 5, one 7 and one 13. number is ______. The fact that “Every composite number can be written uniquely as the product of power of primes” is called Fundamental Theorem of Arithmetic. 2.4.1 Significance of the Fundamental Theorem of Arithmetic The fundamental theorem about natural numbers except 1, that we have stated above has several applications, both in Mathematics and in other fields. The theorem is vastly important in Mathematics, since it highlights the fact that prime numbers are the ‘Building Blocks’ for all the positive integers. Thus, prime numbers can be compared to atoms making up a molecule. Numbers and Sequences 45 X Std_EM Final.indb 45 05-03-2019 17:58:19  1. If a prime number p divides ab then either p divides a or p divides b. That is p divides at least one of them. 2. If a composite number n divides ab, then n neighter divide a nor b. For example, 6 divides 4 × 3 but 6 neither divide 4 nor 3. m Example 2.7 In the given factor tree, find the numbers m and n. Solution Value of the first box from bottom = 5 × 2 = 10 2 Value of n = 5 × 10 = 50 Value of the second box from bottom = 3 × 50 = 150 3 n Value of m = 2 × 150 = 300 5 Thus, the required numbers are m = 300, n = 50 Example 2.8 Can the number 6n , n being a natural number end with 5 2 Fig.2.3 the digit 5? Give reason for your answer. Solution Since 6n = (2 × 3)n = 2n × 3n , Progress Check 2 is a factor of 6n. So, 6n is always even. 1. Let m divides n. Then G.C.D. But any number whose last digit is 5 is always odd. and L.C.M. of m, n are ____ Hence, 6n cannot end with the digit 5. and ____. Example 2.9 Is 7 × 5 × 3 × 2 + 3 a composite 2. The H.C.F. of numbers of the number? Justify your answer. form 2m and 3n is _____. Solution Yes, the given number is a composite number, because 7 × 5 × 3 × 2 + 3 = 3 × (7 × 5 × 2 + 1) = 3 × 71 Since the given number can be factorized in terms of two primes, it is a composite number. Example 2.10 ‘a’ and ‘b’ are two positive integers such that a b × ba = 800. Find ‘a’ and ‘b’. Solution The number 800 can be factorized as Thinking Corner 5 800 = 2 × 2 × 2 × 2 × 2 × 5 × 5 = 2 × 5 2 Can you think of positive Hence, a b × ba = 25 × 52 integers a, b such that a b = ba ? This implies that a = 2 and b = 5 (or) a = 5 and b = 2. Activity 3 Can you find the 4-digit pin number ‘pqrs’ of an ATM card such that p 2 × q 1 × r 4 × s 3 =3, 15, 000 ? Fig.2.4 46 10th Standard Mathematics X Std_EM Final.indb 46 05-03-2019 17:58:22  Exercise 2.2 1. For what values of natural number n, 4n can end with the digit 6? 2. If m, n are natural numbers, for what values of m, does 2n ´ 5m ends in 5? 3. Find the H.C.F. of 252525 and 363636. 4. If 13824 = 2a × 3b then find a and b. 5. If p1x × p2x × p3x × p4x = 113400 where p1, p2 , p3 , p4 are primes in ascending order and 1 2 3 4 x 1, x 2 , x 3 , x 4 are integers, find the value of p1, p2 , p3 , p4 and x 1, x 2 , x 3 , x 4. 6. Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of arithmetic. 7. Find the greatest number consisting of 6 digits which is exactly divisible by 24,15,36? 8. What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case? 9. Find the least number that is divisible by the first ten natural numbers. 2.5 Modular Arithmetic In a clock, we use the numbers 1 to 12 to represent the time period of 24 hours. How is it possible to represent the 24 hours of a day in a 12 number format? We use 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 and after 12, we use 1 instead of 13 and 2 instead of 14 and so on. That is after 12 we again start from 1, 2, 3,... In this system the numbers wrap around 1 to 12. This type of wrapping around after hitting some value is called Modular Arithmetic. Fig.2.5 In Mathematics, modular arithmetic is a system of arithmetic for integers where numbers wrap around a certain value. Unlike normal arithmetic, Modular Arithmetic process cyclically. The ideas of Modular arithmetic was developed by great German mathematician Carl Friedrich Gauss, who is hailed as the “Prince of mathematicians”. Examples 1. The day and night change repeatedly. Life Cycle of Plant 2. The days of a week occur cyclically from Sunday to Saturday. 3. The life cycle of a plant. 4. The seasons of a year change cyclically. (Summer, Autumn, Winter, Spring) 5. The railway and aeroplane timings also work cyclically. The railway time starts at 00:00 and continue. After reaching 23:59, the next minute will become 00:00 instead of 24:00. Fig.2.6 Numbers and Sequences 47 X Std_EM Final.indb 47 05-03-2019 17:58:23  2.5.1 Congruence Modulo Two integers a and b are congruence modulo n if they differ by an integer multiple of n. That b − a = kn for some integer k. This can also be written as a º b (mod n). Here the number n is called modulus. In other words, a º b (mod n) means a - b is divisible by n. For example, 61 º 5 (mod 7) because 61 – 5 = 56 is divisible by 7. Note 1. When a positive integer is divided by n, then the possible remainders are 0, 1, 2,... , n -1. 2. Thus, when we work with modulo n, we replace all the numbers by their remainders upon division by n, given by 0,1,2,3,..., n - 1. Two illustrations are provided to understand modulo concept more clearly. 8 Illustration 1 To find 8 (mod 4) 4 With a modulus of 4 (since the possible remainders are 0, 1, 2, 3) we make a diagram like a clock with numbers 0,1,2,3. 0 We start at 0 and go through 8 numbers in a clockwise sequence 7 3 3 1 1 5 2 1, 2, 3, 0, 1, 2, 3, 0. After doing so cyclically, we end at 0. 2 Therefore, 8 º 0 (mod 4) 6 Illustration 2 Fig.2.7 To find -5 (mod 3) 3 With a modulus of 3 (since the possible remainders are 0, 1, 2, 3) we make a diagram like a clock with numbers 0, 1, 2. 0 2 1 We start at 0 and go through 5 numbers in anti-clockwise 1 2 5 sequence 2, 1, 0, 2, 1. After doing so cyclically, we end at 1. 4 Therefore, −5 ≡ 1 (mod 3) Fig.2.8 2.5.2 Connecting Euclid’s Division lemma and Modular Arithmetic Let m and n be integers, where m is positive. Then by Euclid’s division lemma, we can write n = mq + r where 0 ≤ r < m and q is an integer. Instead of writing n = mq + r we can use the congruence notation in the following way. We say that n is congruent to r modulo m, if n = mq + r for some integer q. Progress Check n = mq + r 1. Two integers a and b are congruent n–r = mq modulo n if ___________. 2. The set of all positive integers which n–r º 0 (mod m) leave remainder 5 when divided by n º r (mod m) 7 are ___________. Thus the equation n = mq + r through Euclid’s Division lemma can also be written as n º r (mod m). 48 10th Standard Mathematics X Std_EM Final.indb 48 05-03-2019 17:58:25  Note Two integers a and b are congruent modulo Thinking Corner m, written as a º b (mod m), if they leave How many integers exist which leave the same remainder when divided by m. a remainder of 2 when divided by 3? 2.5.3 Modulo operations Similar to basic arithmetic operations like addition, subtraction and multiplication performed on numbers we can think of performing same operations in modulo arithmetic. The following theorem provides the information of doing this. Theorem 5 a, b, c and d are integers and m is a positive integer such that if a º b (mod m) and c º d (mod m) then (i) (a + c) ≡ (b + d ) (mod m) (ii) (a − c) ≡ (b − d ) (mod m) (iii) (a × c) ≡ (b × d ) (mod m) Illustration 3 If 17 º 4 (mod 13) and 42 º 3 (mod 13) then from theorem 5, (i) 17 + 42 ≡ 4 + 3 (mod 13) 59 º 7 (mod 13) (ii) 17 - 42 ≡ 4 − 3 (mod 13) -25 º 1 (mod 13) (iii) 17 ´ 42 ≡ 4 × 3 (mod 13) 714 º 12 (mod 13) Theorem 6 If a º b (mod m) then (i) ac º bc (mod m) (ii) a ± c º b ± c (mod m) for any integer c Progress Check 1. The positive values of k such that (k − 3) ≡ 5 (mod11) are _________. 2. If 59 º3 (mod 7), 46 º4 (mod 7) then 105 º _______ (mod 7), 13 º _______ (mod 7), 413 º _______ (mod 7), 368 ≡ _______ (mod 7). 3. The remainder when 7 ´ 13 ´ 19 ´ 23 ´ 29 ´ 31 is divided by 6 is ________. Numbers and Sequences 49 X Std_EM Final.indb 49 05-03-2019 17:58:29  Example 2.11 Find the remainders when 70004 and 778 is divided by 7. Solution Since 70000 is divisible by 7 70000 º 0 (mod 7) 70000 + 4 ≡ 0 + 4 (mod 7) 70004 º 4 (mod 7) Therefore, the remainder when 70004 is divided by 7 is 4. Since 777 is divisible by 7 777 º 0 (mod 7) 777 + 1 ≡ 0 + 1 (mod 7) 778 º 1 (mod 7) Therefore, the remainder when 778 is divided by 7 is 1. Example 2.12 Determine the value of d such that 15 º 3 (mod d). Solution 15 º 3 (mod d) means 15 − 3 = kd, for some integer k. 12 = kd. gives d divides 12. The divisors of 12 are 1,2,3,4,6,12. But d should be larger than 3 and so the possible values for d are 4,6,12. Example 2.13 Find the least positive value of x such that (i) 67 + x ≡ 1 (mod 4) (ii) 98 ≡ (x + 4) (mod 5) Solution (i) 67 + x º 1 (mod 4) 67 + x − 1 = 4n , for some integer n 66 + x = 4n 66 + x is a multiple of 4. Therefore, the least positive value of x must be 2, since 68 is the nearest multiple of 4 more than 66. (ii) 98 ≡ (x + 4) (mod 5) 98 − (x + 4) = 5n , for some integer n. 94 - x = 5n 94 - x is a multiple of 5. Therefore, the least positive value of x must be 4 Since 94 − 4 = 90 is the nearest multiple of 5 less than 94. 50 10th Standard Mathematics X Std_EM Final.indb 50 05-03-2019 17:58:31  Note While solving congruent equations, we get infinitely many solutions compared to finite number of solutions in solving a polynomial equation in Algebra. Example 2.14 Solve 8x º 1 (mod 11) Solution 8x º 1 (mod 11) can be written as 8x − 1 = 11k, for some integer k. 11k + 1 x= 8 When we put k = 5, 13, 21, 29,... then 11k+1 is divisible by 8. 11 × 5 + 1 x= =7 8 11 × 13 + 1 x= = 18 8 Therefore, the solutions are 7,18,29,40, … Example 2.15 Compute x, such that 104 º x (mod 19) Solution 102 = 100 ≡ 5 (mod 19) 104 = (102 )2 ≡ 52 (mod 19) 104 º 25 (mod 19) 104 º 6 (mod 19) (since 25 ≡ 6 (mod 19)) Therefore, x = 6. Example 2.16 Find the number of integer solutions of 3x º 1 (mod 15). Solution 3x º 1 (mod 15) can be written as 3x − 1 = 15k for some integer k 3x = 15k + 1 15k + 1 x= 3 1 x = 5k + 3 1 Since 5k is an integer, 5k + cannot be an integer. 3 So there is no integer solution. Example 2.17 A man starts his journey from Chennai to Delhi by train. He starts at 22.30 hours on Wednesday. If it takes 32 hours of travelling time and assuming that the train is not late, when will he reach Delhi? Numbers and Sequences 51 X Std_EM Final.indb 51 05-03-2019 17:58:34  Solution Starting time 22.30, Travelling time 32 hours. Here we use modulo 24. The reaching time is 22.30+32 (mod 24) º 54.30 (mod24) º6.30 (mod24) (Since 32 = (1×24) + 8 Thursday Friday) Thus, he will reach Delhi on Friday at 6.30 hours. Example 2.18 Kala and Vani are friends. Kala says, “Today is my birthday” and she asks Vani, “When will you celebrate your birthday?” Vani replies, “Today is Monday and I celebrated my birthday 75 days ago”. Find the day when Vani celebrated her birthday. Solution Let us associate the numbers 0, 1, 2, 3, 4, 5, 6 to represent the weekdays from Sunday to Saturday respectively. Vani says today is Monday. So the number for Monday is 1. Since Vani’s birthday was 75 days ago, we have to subtract 75 from 1 and take the modulo 7, since a week contain 7 days. –74 (mod 7) º –4 (mod 7) º 7–4 (mod 7) º 3 (mod 7) (Since, −74 – 3 = −77 is divisible by 7) Thus, 1 − 75 ≡ 3 (mod 7) The day for the number 3 is Wednesday. Therefore, Vani’s birthday must be on Wednesday. Exercise 2.3 1. Find the least positive value of x such that (i) 71 º x (mod 8) (ii) 78 + x ≡ 3 (mod 5) (iii) 89 ≡ (x + 3) (mod 4) x (iv) 96 º (mod 5) (v) 5x º 4 (mod 6) 7 2. If x is congruent to 13 modulo 17 then 7x - 3 is congruent to which number modulo 17? 3. Solve 5x º 4 (mod 6) 4. Solve 3x − 2 ≡ 0 (mod 11) 5. What is the time 100 hours after 7 a.m.? 6. What is the time 15 hours before 11 p.m.? 7. Today is Tuesday. My uncle will come after 45 days. In which day my uncle will be coming? 8. Prove that 2n + 6 × 9n is always divisible by 7 for any positive integer n. 9. Find the remainder when 281 is divided by 17. 52 10th Standard Mathematics X Std_EM Final.indb 52 05-03-2019 17:58:35  10. The duration of flight travel from Chennai to London through British Airlines is approximately 11 hours. The airplane begins its journey on Sunday at 23:30 hours. If the time at Chennai is four and half hours ahead to that of London’s time, then find the time at London, when will the flight lands at London Airport. 2.6 Sequences Consider the following pictures. There is some pattern or arrangement in these 0.5 ft. pictures. In the first picture, 0.5 ft. the first row contains one 0.5 ft. apple, the second row 0.5 ft. contains two apples and in Fig.2.9 the third row there are three apples etc... The number of apples in each of the rows are 1, 2, 3,... In the second picture each step have 0.5 feet height. The total height of the steps from the base are 0.5 feet,1 feet, 1.5 feet,... In the third picture one square, 3 squares, 5 squares,... These numbers belong to category called “Sequences”. Definition A real valued sequence is a function defined on the set of natural numbers and taking real values. Each element in the sequence is called a term of the sequence. The element in the first position is called the first term of the sequence. The element in the second position is called second term of the sequence and so on. If the n th term is denoted by an , then a1 is the first term, a2 is the second term, and so on. A sequence can be written as a1, a2 , a 3 ,..., an ,... Illustration 1. 1,3,5,7,... is a sequence with general term an = 2n − 1. When we put n = 1, 2, 3,..., we get a1 =1, a2 = 3, a 3 = 5, a 4 = 7,... 1 1 1 1 1 2. , , , ,... is a sequence with general term. When we put n = 1,2,3,.... we get 2 3 4 5 n +1 1 1 1 1 a1 = , a2 = , a 3 = , a 4 = ,... 2 3 4 5 If the number of elements in a sequence is finite then it is called a Finite sequence. If the number of elements in a sequence is infinite then it is called an Infinite sequence. Numbers and Sequences 53 X Std_EM Final.indb 53 05-03-2019 17:58:37  Sequence as a Function N f R 1 a1 A sequence can be considered as a function defined on the 2 a2 set of natural numbers N. In particular, a sequence is a function 3 a3 f : N ® R , where R is the set of all real numbers. 4 a4 5 a5 If the sequence is of the form a1, a2 , a 3 ,... then we can associate   the function to the sequence a1, a2 , a 3 ,... by f (k ) = ak , k = 1,2,3,... Progress Check 1. Fill in the blanks for the following sequences (i) 7, 13, 19, _____ ,... (ii) 2, _____, 10, 17, 26,… (iii) 1000, 100, 10, 1, _____,... 2. A sequence is a function defined on the set of _____. 3. The nth term of the sequence 0,2,6,12,20,... can be expressed as _____. 4. Say True or False (i) All sequences are functions (ii) All functions are sequences. Example 2.19 Find the next three terms of the sequences 1 1 1 (i) , , ,.... (ii) 5, 2, -1, -4,.... (iii) 1, 0.1, 0.01,... 2 6 14 1 1 1 1 Solution (i) , , , , 2 6 10 14... +4 +4 +4 In the above sequence the numerators are same and the denominator is increased by 4. 1 1 So the next three terms are a 5 = = 14 + 4 18 Note 1 1 Though all the sequences = a6 = 18 + 4 22 are functions, not all the 1 1 functions are sequences. a 7 = = 22 + 4 26 (ii) 5, 2, –1, –4,... –3 –3 –3 Here each term is decreased by 3. So the next three terms are -7, -10, -13. (iii) 1, 0.1, 0.01,... ÷ 10 ÷ 10 54 10th Standard Mathematics X Std_EM Final.indb 54 05-03-2019 17:58:39  Here each term is divided by 10. Hence, the next three terms are 0.01 a4 = = 0.001 10 0.001 a5 = = 0.0001 10 0.0001 a6 = = 0.00001 10 Example 2.20 Find the general term for the following sequences 1 2 3 (i) 3, 6, 9,... (ii) , , ,... (iii) 5, -25, 125,... 2 3 4 Solution (i) 3, 6, 9,... Here the terms are multiples of 3. So the general term is an = 3n, 1 2 3 (ii) , , ,... 2 3 4 1 2 3 a1 = ; a 2 = ; a 3 = 2 3 4 We see that the numerator of nth term is n, and the denominator is one n more than the numerator. Hence, an = ,n ∈ N n +1 (iii) 5, -25, 125,... The terms of the sequence have + and – sign alternatively and also they are in powers of 5. So the general term an = (−1)n +1 5n , n ∈ N Example 2.21 The general term of a sequence is defined as n (n + 3); n ∈ N is odd an =  n 2 + 1 ; n ∈ N is even  Find the eleventh and eighteenth terms. Solution To find a11 , since 11 is odd, we put n = 11 in an = n(n + 3) Thus, the eleventh term a11 =11(11 + 3) = 154. To find a18 , since 18 is even, we put n = 18 in an = n 2 + 1 Thus, the eighteenth term a18 = 182 + 1 = 325. Example 2.22 Find the first five terms of the following sequence. an −1 a1 = 1 , a 2 = 1 , a n = ; n ≥ 3, n ∈ N an −2 + 3 Solution The first two terms of this sequence are given by a1 = 1 , a2 = 1. The third term a3 depends on the first and second terms. Numbers and Sequences 55 X Std_EM Final.indb 55 05-03-2019 17:58:42  a 3−1 a2 1 1 a3 = = = = a 3−2 + 3 a1 + 3 1+3 4 Similarly the fourth term a 4 depends upon a2 and a 3. 1 1 a 4−1 a3 1 1 1 a4 = = = 4 = 4 = × = a 4−2 + 3 a2 + 3 1 + 3 4 4 4 16 In the same way, the fifth term a 5 can be calculated as 1 a 5−1 a4 16 1 4 1 a5 = = = = × = a 5−2 + 3 a3 + 3 1 16 13 52 +3 4 1 1 1 Therefore, the first five terms of the sequence are 1, 1, , , 4 16 25 Exercise 2.4 1. Find the next three terms of the following sequence. 1 2 3 (i) 8, 24, 72, … (ii) 5, 1, -3, … , , ,… (iii) 4 9 16 2. Find the first four terms of the sequences whose nth terms are given by (i) an = n 3 − 2 (ii) an = (−1)n +1 n(n + 1) (iii) an = 2n 2 − 6 3. Find the n th term of the following sequences 1 2 (i) 2, 5, 10, 17,... (ii) 0, , ,... (iii) 3, 8, 13, 18,... 2 3 4. Find the indicated terms of the sequences whose n th terms are given by 5n (i) an = ; a6 and a13 (ii) an = − (n 2 − 4); a 4 and a11 n +2 n 2 − 1  ; n is even, n ∈ N th n + 3  5. Find a8 and a15 whose n term is an =  2  n  ; n is odd, n ∈ N  2n + 1 6. If a1 = 1, a2 = 1 and an = 2 an −1 + an −2, n ≥ 3, n ∈  , then find the first six terms of the sequence. 2.7 Arithmetic Progression Let us begin with the following two illustrations. Illustration 1 Make the following figures using match sticks (i) How many match sticks are required for each figure? 3,5,7 and 9. Fig.2.11 (ii) Can we find the difference between the successive numbers? 5−3 = 7−5 = 9−7 = 2 Therefore, the difference between successive numbers is always 2. 56 10th Standard Mathematics X Std_EM Final.indb 56 05-03-2019 17:58:46  Illustration 2 A man got a job whose initial monthly salary is fixed at ₹10,000 with an annual increment of ₹ 2000. His salary during 1st , 2nd and 3rd years will be ₹ 10000 , ₹ 12000 and ₹ 14000 respectively. If we now calculate the difference of the salaries for the successive years, we get 12000 – 10000 = 2000; 14000 – 12000 = 2000. Thus the difference between the successive numbers (salaries) is always 2000. Did you observe the common property behind these two illustrations? In these two examples, the difference between successive terms always remains constant. Moreover, each term is obtained by adding a fixed number (2 and 2000 in illustrations 1 and 2 presented above) to the preceding term except the first term. This fixed number which is a constant for the differences between successive terms is called the “common difference”. Definition Let a and d be real numbers. Then the numbers of the form a, a + d , a + 2d , a + 3d , a + 4d ,... is said to form Arithmetic Progression denoted by A.P. The number ‘a’ is called the first term and ‘d’ is called the common difference. Simply, an Arithmetic Progression is a sequence whose successive terms differ by a constant number. Thus, for example, the set of even positive integers 2, 4, 6, 8, 10, 12,… is an A.P. whose first term is a = 2 and common difference is also d = 2 since 4 − 2 = 2, 6 − 4 = 2, 8 − 6 = 2, … Most of common real−life situations often produce numbers in A.P. Note 1. The difference between any two consecutive terms of an A.P is always constant. That constant value is called the common difference. 2. If there are finite numbers of terms in an A.P. then it is called Finite Arithmetic Progression. If there are infinitely many terms in an A.P then it is called Infinite Arithmetic Progression. 2.7.1 Terms and Common Difference of an A.P. 1. The terms of an A.P. can be written as t1 = a = a + (1 − 1)d , t2 = a + d = a + (2 − 1)d , t3 = a + 2d = a + (3 − 1)d , t4 = a + 3d = a + (4 − 1)d ,... In general, the nth term denoted by tn can be written as tn = a + (n − 1)d. 2. In general to find the common difference of an A.P. we should subtract first term from the second term, second from the third and so on. Numbers and Sequences 57 X Std_EM Final.indb 57 05-03-2019 17:58:48  For example, t1 = a, t2 = a + d Therefore, t2 − t1 = (a + d ) − a = d Similarly, t2 = a + d, t3 = a + 2d,... Therefore, t3 − t2 = (a + 2d ) − (a + d ) = d In general, d = t2 − t1 = t3 − t2 = t4 − t3 =.... So, d = tn − tn −1 for n = 2, 3, 4,... Progress Check 1. The difference between any two consecutive terms of an A.P. is _______. 2. If a and d are the first term and common difference of an A.P. then the 8th term is _______. 3. If tn is the n th term of an A.P., then t2n -tn is _______. Let us try to find the common differences The common difference of of the following A.P.’s an A.P. can be positive, negative or zero. (i) 1, 4, 7, 10,¼ d = 4 − 1 = 7 − 4 = 10 − 7 =... = 3 Thinking Corner (ii) 6, 2, −2, −6,… If tn is the n th term of an A.P. then d = 2 − 6 = −2 − 2 = −6 − (−2) =... = −4 the value of tn +1 −tn −1  is _______. Example 2.23 Check whether the following sequences are in A.P or not? (i) x + 2, 2x + 3, 3x + 4,... (ii) 2, 4, 8, 16,... (iii) 3 2, 5 2, 7 2, 9 2,... Solution To check that the given sequence is in A.P., it is enough to check if the differences between the consecutive terms are equal or not. (i) t2 - t1 = (2x + 3) − (x + 2) = x + 1 t3 - t2 = (3x + 4) − (2x + 3) = x + 1 t2 - t1 = t3 − t2 Thus, the differences between consecutive terms are equal. Hence the sequence x + 2, 2x + 3, 3x + 4,... is in A.P. (ii) t2 - t1 = 4 − 2 = 2 t3 - t2 = 8 − 4 = 4 t2 - t1 = t3 − t2 Thus, the differences between consecutive terms are not equal. Hence the terms of the sequence 2, 4, 8, 16,... are not in A.P. 58 10th Standard Mathematics X Std_EM Final.indb 58 05-03-2019 17:58:52  (iii) t2 - t1 = 5 2 − 3 2 = 2 2 t 3 - t2 = 7 2 − 5 2 = 2 2 t4 - t3 = 9 2 − 7 2 = 2 2 Thus, the differences between consecutive terms are equal. Hence the terms of the sequence 3 2, 5 2, 7 2, 9 2,... are in A.P. Example 2.24 Write an A.P. whose first term is 20 and common difference is 8. Solution First term = a = 20 ; common difference = d= 8 Arithmetic Progression is a, a + d, a + 2d, a + 3d,... In this case, we get 20, 20 + 8, 20 + 2(8), 20 + 3(8),... So, the required A.P. is 20, 28, 36, 44, … Note An Arithmetic progression having a common difference of zero is called a constant arithmetic progression. Activity 5 There are five boxes here. You have to pick one number from each box and form five Arithmetic Progressions. 8 6 14 1 26 3.5 20 11 32 8.5 –8 40 –14 –11 –2 12 85 –5 2 7 55 –3 70 17 100 Example 2.25 Find the 15th , 24th and n th term (general term) of an A.P given by 3, 15, 27, 39,… Solution We have, first term = a = 3 and common difference = d = 15 − 3 = 12. We know that n th term (general term) of an A.P with first term a and common difference d is given by tn = a + (n − 1)d t15  =a +(15 − 1)d = a + 14d = 3 + 14 (12) = 171 (Here a = 3 and d = 12) t24 = a + (24 − 1)d =a +23d =3 +23(12) = 279 The n th (general term) term is given by tn = a + (n − 1)d Thus, tn = 3 + (n − 1)12 tn = 12n − 9 Numbers and Sequences 59 X Std_EM Final.indb 59 05-03-2019 17:58:56  Note In a finite A.P. whose first term is a and last term l, then the number of terms in the l − a  A.P. is given by l = a + (n − 1)d gives n =   + 1  d  Example 2.26 Find the number of terms in the A.P. 3, 6, 9, 12,…, 111. Solution Progress Check First term a = 3 ; common difference d = 6 − 3 = 3 ; last term l = 111 1. The common difference of a l − a  constant A.P. is _______. We know that, n =   + 1 2. If a and l are first and last terms of  d  an A.P. then the number of terms is 111 − 3  n =   + 1 = 37 _______.  3  Thus the A.P. contain 37 terms. Example 2.27 Determine the general term of an A.P. whose 7th term is −1 and 16th term is 17. Solution Let the A.P be t1, t2 , t3 , t4 ,... It is given that t7 = −1 and t16 = 17 a + (7 − 1)d = −1 and a + (16 − 1)d = 17 a + 6d = −1...(1) a + 15d = 17...(2) Subtracting equation (1) from equation (2), we get 9d = 18 gives d = 2 Putting d = 2 in equation (1), we get a + 12 = −1 so a = –13 Hence, general term tn = a + (n − 1)d = −13 + (n − 1) × 2 = 2n − 15 Example 2.28 If l th , m th and nth terms of an A.P. are x, y, z respectively, then show that (i) x (m − n ) + y (n − l ) + z (l − m ) = 0 (ii) (x − y )n + (y − z )l + (z− x )m = 0 Solution (i) Let a be the first term and d be the common difference. It is given that tl = x , tm = y, tn = z Using the general term formula a + (l − 1)d = x...(1) a + (m − 1)d = y...(2) a + (n −1)d = z...(3) 60 10th Standard Mathematics X Std_EM Final.indb 60 05-03-2019 17:59:00  We have, x (m − n ) + y(n − l ) + z (l − m ) a[(m − n) + (n − l ) + (l − m)] + d[(m − n)(l − 1) + (n − l )(m − 1) + (l − m)(n − 1)] = a + d[lm − ln − m + n + mn − lm − n + l + ln − mn − l + m ] = a(0) + d(0) = 0 (ii) On subtracting equation (2) from equation (1), equation (3) from equation (2) and equation (1) from equation (3), we get x − y = (l − m )d y − z = (m − n )d z − x = (n − l )d (x − y )n + (y − z )l + (z − x )m = [(l − m )n + (m − n )l + (n − l )m ]d = ln − mn + lm − nl + nm − lm  d = 0 Note In an Arithmetic Progression (i) If every term is added or subtracted by a constant, then the resulting sequence is also an A.P. (ii) If every term is multiplied or divided by a non-zero number, then the resulting sequence is also an A.P. (iii) If the sum of three consecutive terms of an A.P. is given, then they can be taken as a - d, a and a + d. Here the common difference is d. (iv) If the sum of four consecutive terms of an A.P. is given then, they can be taken as a - 3d , a - d , a + d and a + 3d. Here common difference is 2d. Example 2.29 In an A.P., sum of four consecutive terms is 28 and their sum of their squares is 276. Find the four numbers. Solution Let us take the four terms in the form (a - 3d ), (a - d ), (a + d ) and (a + 3d ). Since sum of the four terms is 28, a − 3d + a − d + a + d + a + 3d = 28 4a = 28 gives a = 7 Similarly, since sum of their squares is 276, (a − 3d )2 + (a − d )2 + (a + d )2 + (a + 3d )2 = 276. a 2 − 6ad + 9d 2 + a 2 − 2ad + d 2 + a 2 + 2ad + d 2 + a 2 + 6ad + 9d 2 = 276 4a 2 +20d 2 =276 ⇒ 4(7)2 + 20d 2 = 276. d 2 = 4 gives d = ± 2 If d = 2 then the four numbers are 7 - 3(2), 7 - 2, 7 + 2, 7+3(2) That is the four numbers are 1, 5, 9 and 13. Numbers and Sequences 61 X Std_EM Final.indb 61 05-03-2019 17:59:04  If a = 7, d = −2 then the four numbers are 13, 9, 5 and 1 Therefore, the four consecutive terms of the A.P. are 1, 5, 9 and 13. Condition for three numbers to be in A.P If a, b, c are in A.P. then b = a + d, c = a + 2d so a + c = 2a + 2d 2(a + d ) = 2b Thus 2b 2b = a + c Similarly, if 2b = a + c, then b − a = c − b so a, b, c are in A.P. Thus three non-zero numbers a, b, c are in A.P. if and only if 2b = a + c Example 2.30 A mother devides ₹207 into three parts such that the amount are in A.P. and gives it to her three children. The product of the two least amounts that the children had ₹4623. Find the amount received by each child. Solution Let the amount received by the three children be in the form of A.P. is given by a - d , a, a + d. Since, sum of the amount is ₹207, we have (a − d ) + a + (a + d ) = 207 3a = 207 gives a = 69 It is given that product of the two least amounts = 4623. (a − d )a = 4623 (69 − d )69 = 4623 d =2 Therefore, amount given by the mother to her three children are ₹(69−2), ₹69, ₹(69+2). That is, ₹67, ₹69 and ₹71. Progress Check 1. If every term of an A.P. is multiplied by 3, then the common difference of the new A.P. is _______. 2. Three numbers a, b and c will be in A.P. if and only if _______. Exercise 2.5 1. Check whether the following sequences are in A.P. 1 1 1 1 (i) a - 3, a - 5, a - 7,... (ii) , , , ,... (iii) 9, 13, 17, 21, 25,... 2 3 4 5 -1 1 2 (iv) , 0, , ,... (v) 1, -1, 1, -1, 1, -1,... 3 3 3 2. First term a and common difference d are given below. Find the corresponding A.P. 62 10th Standard Mathematics X Std_EM Final.indb 62 05-03-2019 17:59:06  3 1 (i) a = 5 , d = 6 (ii) a = 7 , d = −5 (iii) a = ,d= 4 2 3. Find the first term and common difference of the Arithmetic Progressions whose nth terms are given below (i) tn = −3 + 2n (ii) tn = 4 − 7n 4. Find the 19th term of an A.P. -11, -15, -19,... 5. Which term of an A.P. 16, 11, 6, 1,... is -54 ? 6. Find the middle term(s) of an A.P. 9, 15, 21, 27,…,183. 7. If nine times ninth term is equal to the fifteen times fifteenth term, show that six times twenty fourth term is zero. 8. If 3 + k, 18 - k, 5k + 1 are in A.P. then find k. 9. Find x, y and z, given that the numbers x, 10, y, 24, z are in A.P. 10. In a theatre, there are 20 seats in the front row and 30 rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row? 11. The sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms. 12. The ratio of 6th and 8th term of an A.P is 7:9. Find the ratio of 9th term to 13th term. 13. In a winter season let us take the temperature of Ooty from Monday to Friday to be in A.P. The sum of temperatures from Monday to Wednesday is 0° C and the sum of the temperatures from Wednesday to Friday is 18° C. Find the temperature on each of the five days. 14. Priya earned ₹15,000 in the first month. Thereafter her salary increased by ₹1500 per year. Her expenses are ₹13,000 during the first year and the expenses increases by ₹900 per year. How long will it take for her to save ₹20,000 per month. 2.8 Series The sum of the terms of a sequence is called series. Let a1, a2, a 3,..., an ,... be the sequence of real numbers. Then the real number a1 + a2 + a3 +  is defined as the series of real numbers. If a series has finite number of terms then it is called a Finite series. If a series has infinite number of terms then it is called an Infinite series. Let us focus our attention only on studying finite series. Numbers and Sequences 63 X Std_EM Final.indb 63 05-03-2019 17:59:08  2.8.1 Sum to n terms of an A.P. A series whose terms are in Arithmetic progression is called Arithmetic series. Let a, a + d, a + 2d, a + 3d,... be the Arithmetic Progression. The sum of first n terms of a Arithmetic Progression denoted by Sn is given by, Sn = a + (a + d ) + (a + 2d ) +  + (a + (n − 1)d )...(1) Rewriting the above in reverse order Sn = (a + (n − 1)d ) + (a + (n − 2)d ) +  + (a + d ) + a...(2) Adding (1) and (2) we get, 2Sn =[a +a +(n − 1)d ]+[a +d +a +(n − 2)d ] + + [a +(n −2)d +(a +d )]+[a +(n − 1)d + a ] = [2a + (n − 1)d ] + [2a + (n − 1)d +  + [2a + (n − 1)d ] (n terms) n 2Sn = n × [2a + (n − 1)d ] gives Sn = 2a + (n − 1)d  2   Note If the first term a, and the last term l ( n th term) are given then n n Sn = 2a + (n − 1)d  = a + a + (n − 1)d  since, l = a + (n − 1)d we have   2  2 n Sn = [a + l ]. 2 Progress Check 1. The sum of terms of a sequence is called ______. 2. If a series have finite number of terms then it is called ______. 3. A series whose terms are in ______ is called Arithmetic series. 4. If the first and last terms of an A.P. are given, then the formula to find the sum is ______. 1 1 3 Example 2.31 Find the sum of first 15 terms of the A. P. 8, 7 , 6 , 5 ,... 4 2 4 1 3 Solution Here the first term a = 8, common difference d = 7 −8 =− , 4 4 n Sum of first n terms of an A.P. Sn = 2a + (n − 1)d  2  15  3  S15 = 2 × 8 + (15 − 1)(− ) 2  4  15  21  165 S15 = 16 −  = 2  2  4 64 10th Standard Mathematics X Std_EM Final.indb 64 05-03-2019 17:59:12  Example 2.32 Find the sum of 0.40 + 0.43 + 0.46 +  + 1. Solution Here the value of n is not given. But the last term is given. From this, we can find the value of n. Given a = 0.40 and l = 1 , we find d = 0.43 − 0.40 = 0.03. l − a  Therefore, n =   + 1   d 1 − 0.40  =   + 1 = 21  0.03  n Sum of first n terms of an A.P. Sn = a + l  2 21 Here, n = 21. Therefore, S 21 = 0.40 + 1 = 14.7 2 So, the sum of 21 terms of the given series is 14.7. Example 2.33 How many terms of the series 1 + 5 + 9 +... must be taken so that their sum is 190? Solution Here we have to find the value of n, such that Sn = 190. First term a = 1, common difference d = 5 − 1 = 4. Sum of first n terms of an A.P. n Thinking Corner Sn = [2a + (n − 1)d ] = 190 2 The value of n must be n [2 × 1 + (n − 1) × 4 ] = 190 positive. Why? 2 n[4n - 2] = 380 2n 2 - n - 190 = 0 (n − 10)(2n + 19) = 0 19 But n = 10 as n = − is impossible. Therefore, n = 10. 2 Progress Check State True or False. Justify it. 1. The nth term of any A.P. is of the form pn+q where p and q are some constants. 2. The sum to nth term of any A.P. is of the form pn2+qn + r where p, q, r are some constants. Example 2.34 The 13th term of an A.P. is 3 and the sum of first 13 terms is 234. Find the common difference and the sum of first 21 terms. Solution Given the 13th term = 3 so, t13 = a + 12d = 3...(1) Numbers and Sequences 65 X Std_EM Final.indb 65 05-03-2019 17:59:15  13 Sum of first 13 terms = 234 gives S13 = [2a + 12d ] = 234 2 2a + 12d = 36...(2) −5 Solving (1) and (2) we get, a = 33, d = 2 -5 Therefore, common difference is. 2 21   5  21 −  = [66 − 50] = 168. Sum of first 21 terms S 21 = 2 × 33 + (21 − 1) ×  2  2   2 5n 2 3n Example 2.35 In an A.P. the sum of first n terms is +. Find the 17th term. 2 2 Solution The 17th term can be obtained by subtracting the sum of first 16 terms from the sum of first 17 terms 5 × (17)2 3 × 17 1445 51 S17 = + = + = 748 2 2 2 2 5 × (16)2 3 × 16 1280 48 S16 = + = + = 664 2 2 2 2 Now, t17 = S17 − S16 = 748 − 664 = 84 Example 2.36 Find the sum of all natural numbers between 300 and 600 which are divisible by 7. Solution The natural numbers between 300 and 600 which are divisible by 7 are 301, 308, 315, …, 595. The sum of all natural numbers between 300 and 600 is 301 + 308 + 315 +  + 595. The terms of the above series are in A.P. First term a = 301 ; common difference d = 7 ; Last term l = 595. l − a   595 − 301 n =   + 1 =    + 1 = 43  d  7  n 43 Since, Sn = [a + l ] , we have S 57 = [301 + 595]= 19264. 2 2 Example 2.37 A mosaic is designed in the shape of an equilateral triangle, 12 ft on each side. Each tile in the mosaic is in the shape of an equilateral triangle of 12 inch side. The tiles are alternate in colour as shown in the figure. 12 Find the number of tiles of each colour and total number f t. of tiles in the mosaic. Solution Since the mosaic is in the shape of an equilateral triangle of 12 ft, and the tile is in the shape of an equilateral triangle of 12 inch (1 ft), there will be 12 rows in the mosaic. Fig.2.12 66 10th Standard Mathematics X Std_EM Final.indb 66 05-03-2019 17:59:18  From the figure, it is clear that number of white tiles in each row are 1, 2, 3, 4, …, 12 which clearly forms an Arithmetic Progression. Similarly the number of blue tiles in each row are 0, 1, 2, 3, …, 11 which is also an Arithmetic Progression. 12 Number of white tiles = 1 + 2 + 3 +  + 12 = [1 + 12] = 78 2 12 Number of blue tiles = 0 + 1 + 2 + 3 +  + 11 = [0 + 11] = 66 2 The total number of tiles in the mosaic = 78 + 66 = 144 Example 2.38 The houses of a street are numbered from 1 to 49. Senthil’s house is numbered such that the sum of numbers of the houses prior to Senthil’s house is equal to the sum of numbers of the houses following Senthil’s house. Find Senthil’s house number? Solution Let Senthil’s house number be x. It is given that 1 + 2 + 3 +  + (x − 1) = (x + 1) + (x + 2) +  + 49 1 + 2 + 3 +  + (x − 1) = 1 + 2 + 3 +  + 49 − 1 + 2 + 3 +  + x  x −1  49  x  1 + (x − 1) =  1 + 49 − 1 + x  2 2 2 x (x - 1) 49 × 50 x (x + 1) = − 2 2 2 x - x = 2450 − x − x ⇒ 2x 2 = 2450 2 2 x 2 = 1225 gives x = 35 Therefore, Senthil’s house number is 35. Example 2.39 The sum of first n, 2n and 3n terms of an A.P are S1, S 2 and S3 respectively. Prove that S 3 = 3(S 2 − S1 ). Solution If S1, S 2 and S3 are sum of first n, 2n and 3n terms of an A.P. respectively then n  , S = 2n 2a + (2n − 1)d  , S = 3n 2a + (3n − 1)d  S1 = 2a + (n − 1)d 2  2 2   3 2   2n  n Consider, S 2 - S1 =  2a + (2n − 1)d  − 2a + (n − 1)d  2 2 n = [4a + 2(2n − 1)d ] − [2a + (n − 1)d ] 2 n  Thinking Corner S 2 - S1 = × 2a + (3n − 1)d  2  1. What is the sum of first 3n  3(S 2 - S1 ) = 2a + (3n − 1)d  n odd natural numbers? 2  2. What is the sum of first 3(S 2 - S1 ) = S 3 n even natural numbers? Numbers and Sequences 67 X Std_EM Final.indb 67 05-03-2019 17:59:22  Exercise 2.6 1. Find the sum of the following (i) 3,7,11,… up to 40 terms. (ii) 102,97,92,… up to 27 terms. (iii) 6 + 13 + 20 +  + 97 2. How many consecutive odd integers beginning with 5 will sum to 480? 3. Find the sum of first 28 terms of an A.P. whose nth term is 4n - 3. 4. The sum of first n terms of a certain series is given as 2n 2 - 3n. Show that the series is an A.P. 5. The 104th term and 4th term of an A.P are 125 and 0. Find the sum of first 35 terms. 6. Find the sum of all odd positive integers less than 450. 7. Find the sum of all natural numbers between 602 and 902 which are not divisible by 4. 8. Raghu wish to buy a laptop. He can buy it by paying ₹40,000 cash or by giving it in 10 installments as ₹4800 in the first month, ₹4750 in the second month, ₹4700 in the third month and so on. If he pays the money in this fashion, find