10: SERIES AND SEQUENCES Sequence and Series Consider a sequence whose first term is denoted by � A sequence is a set of numbers in which consecutive and whose difference is denoted by �. The first five terms are connected by a definite rule or pattern. The terms are shown in the table below. numbers in this ordered list are called elements or terms. Sequences vary depending on the pattern or T1 T2 T3 T4 T5 rule that exist between consecutive terms. We also � a + d a + 2d a + 3d a + 4d refer to a sequence as a progression. The sum of the terms of a sequence is called a series. To determine a formula for the �$% term, we look at each consecutive term to discern a pattern. We Finite and Infinite Sequences observe that each term has � and the coefficient of d A finite sequence can be specified by a complete list is always one less than the value of n. of its elements. It consists of a countable number of terms. The set of even numbers from 2 to 10 forms The nth term of an AP the finite sequence, {2, 4, 6, 8, 10}. T = a + n −1 d n () An infinite sequence can be specified by an incomplete list of its elements. It has an ‘infinite’ Example 1 number of terms and in listing the terms, we use a set An arithmetic progression is defined by of three dots after the last term in the list to indicate {4, 2, 0, −2,…}. Determine that it has no end. The set of even numbers forms an infinite sequence, {2, 4, 6, 8, …} i. the common difference, d ii. the 406th term The arithmetic series or progression (AP) A series, in which each term is obtained from the Solution preceding one by the addition of a constant quantity, i. In the AP, {4, 2, 0, −2,…}, is called an arithmetic progression (AP). This The first term � = 4 constant quantity is called the common difference. The common difference, d, can be found by subtracting any two consecutive terms in the The sequence order, �) − �)+, 3, 5, 7, 9,… has a common difference of 2. � = 2 − 4 = 0 − 2 = −2 The sequence Hence, � = −2 20, 18, 16, 14,... has a common difference of − 2 ii. Recall: Tann =+()-1 d, where n is the The sequence number of terms. Substituting for �, � and 1 1 1 4, 3 , 3, 2 ,... has a common difference of − �, we have 2 2 2 T 4 406 1 2 ∴ 406 = + ()− ×() − = 4 + 405× −2 Notice that the common difference can be a positive () or negative or even a fractional number. Its value, = 4 − 810 though, is the same for a particular series and cannot be altered for that series. = −806 Notation for an Arithmetic Progression Sum of the first n terms of an AP 1st term = a www.faspassmaths.com Last term = l For an arithmetic progression, we usually use the th symbol Sn to denote the sum of the first n terms. For Number of terms = n The n term = Tn example, S3 denotes the sum of the first 3 terms, The common difference, d = Tn− Tn-1 TTT++. 123 The nth term of an AP Sometimes we are required to calculate a particular For an AP with n terms, we denote � and � as the term in an AP. We can do so once we know the first first term and the common difference. If the last term term and the common difference. If we require the is denoted by l, then, we can calculate �) by the 20th term, we need to list all the terms. The nth term, following series. Note, for convenience, only the first four terms and the last four terms were included. denoted by Tn can be determined by using a formula. Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 1 of 7 �) = � + (� + �) + (� + 2�) + (� + 3�) + ⋯ Example 3 (� − 3�) + (� − 2�) + (� − �) + � Find the sum of all the terms in the A.P 10, 15, 20, …, 1000 Reversing the order of the terms in the above equation we have: Solution �) = � + (� − �) + (� − 2�) + (� − 3�) + ⋯ The first term, a = 10 and d =15- 10= 5 (� + 3�) + (� + 2�) + (� + �) + � The last term, l = 1000 We do not know the number of terms, n. Adding both equations, we notice that each pair of We can determine n by substituting for a, l and d ( ) consecutive terms add to � + � . For example: in the following formula: Adding the first terms from each Sn, we have (� + �). Adding the second terms from each Sn, we have la=+() n-1 d (� + �) + (� − �) = � + �. 1000=+ 10()n -´ 1 5 Therefore, the sum of both sequences will look like: 990= 5()n - 1 n -1= 198 2�) = (� + �) + (� + �) + (� + �) + ⋯ (� + �) + (� + �) + (� + �) n = 199 Since there are � terms in the sequence, The sum of all 199 terms, S199 , can be found using 2� = �(� + �) ) n n � either or . � = (� + �) Saln =+() {21an+()- d} ) 2 2 2 n Using Sal=+() We can replace l from this formula by treating l as n 2 any term. If we want to find the sum of the first n 199 terms then, S199 =+()10 1000 2 T = a + n −1 d n () = 100 495 So, we may take, \la= +() n-1 d OR n n Using Sandn =+{21()-} \Saandn ={ ++()()-1 } 2 2 n 199 S199 =+{210()() 19915-} Sandn =+{21()-} 2 2 = 100495 The sum of the first n terms of an AP n Example 4 Saln =+() An arithmetic progression with first term 8 has 2 101 terms. If the sum of the first three terms is n Sandn =+{21()-} one-third of the sum of the last three terms, find 2 a. the common difference, d b. the sum of the last 48 terms of the series. Example 2 Find the sum ofwww.faspassmaths.com the first 40 terms of the AP Solution { 4, 6, 8, 10, …} a. Let a = 8 , common difference = d, n =101 The sum of the first 3 terms is: Solution We can deduce that, a = 4, d = 6 – 4 = 2, n = 40 n TTT123++=++ a()() ad ++ a233243 d = a + d = + d The sum of the n terms is Sandn =+{21()-} 2 TT99++=+ 100 T 101 ()()() a98 d ++ a 99 d ++ a 100 d 40 TT99++=+ 100 T 101 329724297 a d =+ d S40 =+{2(4) (40- 1)2} =+= 20{8 2(39)} 1720 2 Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 2 of 7 The sum of the first three terms ii The total distance covered after 30 days is the 1 sum of the first 30 terms of the arithmetic = [sum of the last three terms] progression. 3 n (TT99++ 100 T 101 ) Sandn =+{21()-} ()TTT++ = 2 123 3 1 30 ìü1 S30 =+íý2()() 2 30-´ 1 24+= 3dd() 24 + 297 22 3 îþ 16 = 99d − 3d ìü1 =+15íý 4 14 16 1 îþ2 ∴96d = 16 ⇒ d = , d = 96 6 = 277.5 km Example 6 b. Sum of the last 48 terms = SS101- 53 n Find the sum of all the numbers from 1000 to 2000 Recall Sandn =+{21()-} which are divisible by 5. 2 \ Sum of the last 48 terms = Solution 101 ìüìü1 53 1 íýíý28()()+ 1011-- 28()()+ 531- The series is 1000, 1005, 1010, …, 1995, 2000. 26îþîþ26The terms are in AP with the first term, a = 1000 22 The common difference, d =1005- 1000= 5 =1649- 653 33 = 996 Last term, la=+() n-1 d, where n = number of terms. Example 5 \2000=+ 1000()n - 1 5 A marathon runner begins his first day of training by running 2 km. He then increases this distance 1000=()n - 1 5 by ½ km more each day, from the distance he ran n = 201 the previous day. To find the sum of all the terms from 1000 to i. On what day would he first cover 15 km? 2000, we can use any one of the methods below. ii. What is the total distance covered after 30 n Sal201 =+() days? 2 201 =+()1000 2000 Solution 2 i The distances covered on a daily basis form the = 301500 following AP: OR 111 2, 2 , 3, 3 , 4, 4 ,... n 222 Sandn =+{21()-} 2 where the first term a = 2 and the common 1 201 difference d = . =+{2()() 1000 201- 1 5} 2 2 = 301500 We are looking for n such that Tn = 15 . Tan=+ -1 d Since n www.faspassmaths.com() , then 1 The Geometric Progression (GP) 15=+ 2()n -´ 1 2 A series in which each term is obtained from the 30=+ 4n - 1 () preceding one by the multiplication of a constant n = 27 quantity is called a geometric series or progression (GP). This constant quantity is called the common Hence, on the 27th day, he would run a distance of ratio. exactly 15 km. For any GP, we can find the common ratio by dividing any term by the term that came just before. Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 3 of 7 The common ratio Sum of the first n terms of a GP The geometric series {4, 12, 36, 108, …} has a 4´ 3= 12 12´ 3= 36 common ratio of 3, note , , Let us use the symbol Sn to represent the sum of the 36´ 3= 108 and so on. first n terms of a geometric progression. So, the sum OR We may say 12 ÷ 4 = 3 or 36 ÷ 12 = 3 and so on. of the first three terms, S3 is TTT123++. We will now derive the formula for the sum of the n terms of The geometric series {16, 8, 4, 2, …} has a common a GP. ratio of ½, note 16 ×½ = 8, 8 × ½ =4, 4 × ½ = 2, and so on.