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. Let us remember that each term is r times the OR 8 ÷ 16 = ½, 4 ÷ 8 = ½, 2 ÷ 4 = ½ previous term or the ratio of any term to the term The geometric series {4, −8, 16, −32, …} has a before it is always r:1 common ratio of −2, note 4 × −2 = −8, −8 × −2 =16, S = a + ar + ar 2 + ar 3 + ...+ ar n − 1 (1) 16 ×−2 = −32 and so on n OR −8 ÷ 4 = −2, and 16 ÷ −8 = −2 Equation (1) ´r 234 n rSn =+ ar ar + ar + ar ++... ar (2) Notice that the common ratio may take any value. That is, it may be positive or negative or fractional. Equation (1) – Equation (2) � − �� = � + �� + �� + �� + ⋯ + �� − Notation for a Geometric Progression (�� + �� + �� + ⋯ �� + ��) First term = a Number of terms = n Notice all the terms will cancel out except the first th The n term = Tn term of equation (1) and the last term of equation (2) T and which now carries a minus sign before it. Hence, The common ratio, r is n SrSaar-=-n Tn-1 nn n Srarn ()11-=()- The nth term of a Geometric Progression ar()1- n Sr=<1 th n The n term of a GP, denoted by Tn can be 1- r determined by using a formula. Consider a sequence whose first term is denoted by � and whose ratio is Whenr >1 , the subtraction is taken in the reverse denoted by �. The first five terms are shown in the order and produces the alternate form of the formula. table below.

ar()n -1 T1 T2 T3 T4 T5 Sr=>1 2 3 4 n a ar ar ar ar r -1

From the above table, we observe that the coefficient The formula for the sum of n terms in a GP can now of r is a and the power of r is always one less than be stated. the value of n. The sum of the first n terms of a GP The nth term of a G.P n - 1 If the first term of a GP is denoted by a, the Tarn = number of terms by n, the sum of the first n terms www.faspassmaths.com Example 7 by Sn and the common ratio by r, then A geometric progression has terms {4, 8, 16,….} ar()n -1 ar()1- n S = S = Determine the 20th term. n n r -1 1- r r >1 r <1 Solution The first term, a = 4 , the common ratio 8 Example 8 r = = 2, r > 1 . Find the sum of the first 40 terms in the geometric 4 th 20-1 19 progression {4, 8, 16, …}. The 20 term is T20 = 4(2) = 4(2) =2097152

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 4 of 7 Solution Starting with the formula for the sum of � terms of The first term, a = 4 , the common ratio, any converging GP, we have 8 r = = 2, r > 1 n n 4 a()1− r a − ar aarn Sn = = = - The sum of the first n terms is: 1− r 1− r 11--rr As the value of n ®¥ , then, rn gets smaller and n a r −1 smaller, that is, it tends to 0. Hence, as rn ® 0, the S = () n r −1 arn 40 term ® 0 . 4()() 2 −1 1- r S40 = aa 2 −1 Therefore, Sn approaches -0 = . 40 11--rr = 4() 2 −1 () This means that when r <1 , the more and more terms of the GP that we add, the closer and closer we Divergent and Convergent Series a would get to the value of . We say that the 1- r Consider the sum of the terms of an AP in which � = a 3 and � = 2. series converges with limit and this value is 1- r � = 3 + 5 + +7 + 9 + called its sum to infinity.

Notice that the terms of the progression are The sum to infinity of a GP increasing and as � gets larger the sum will approach a infinity. It is impossible to sum all the terms of this S¥ = sequence. This is because this sum approaches 1- r infinity. All AP’s have sums that approach infinity with the first term, a and common ratio, r. and such a series is a divergent series. This exists only when, r <1 .

Consider the sum of the terms of a GP in which Example 9 r >1. For example, let � = 3 and � = 2. Find the sum to infinity of the geometric series

� = 3 + 6 + 12 + 24 + 48 + ⋯ {8, 4, 2,…}

Notice that the terms of the progression are Solution increasing and as � gets larger the sum will approach In the geometric progression {8, 4, 2,…} the first infinity. It is impossible to sum all the terms of this term, a = 8 . The common ratio, sequence. This is because this sum approaches 4 1 infinity. This type of GP is divergent. r = = , r < 1 8 2

a 8 Now, let us consider the sum of the terms of a GP S ===16 ¥ 11--r 1 when r <1 . For example, let � = 243 and � = . 2 1 1 � = 243 + 81 + 27 + 9 + 3 + 1 + + + ⋯ Example 10 3 9 In a geometric progression with the common ratio, Notice that thewww.faspassmaths.com terms of the progression are r, the sum to infinity is four times the first term. decreasing and as � gets larger the terms approach Find the value of r, r <1 . zero. Hence, the sum of the terms approaches a fixed value as we keep adding more and more terms. This Solution is called the sum to infinity. This type of GP is We are given that �= 4 ´ first term convergent. � = � 1 − � Hence, the sum to infinity of a G.P, exists only when 4�(1 − �) = � r <1. We can derive a simple formula to calculate 4(1 − �) = 1 this sum for any GP. 1 − � = 1/4 � = 3/4

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 5 of 7 In some questions, we are given a sequence and the Solution type of sequence is not stated. In such cases, we have to deduce from the data what type of sequence is We calculate the amount of soil removed for each involved. day and look for a pattern.

Example 11 Day 1 2 3 4… An item of jewellery appreciates by 10% of its Amount 6 66 666

of soil 10 10´ 10´´ 10´´´ ... value each year. If its original value is $1 000, find removed 5 55 555 its estimated value at the start of the 11th year. in m3 2 3 6 æö6 æö6 10 10´ 10´ç÷ 10´ç÷ ... 5 5 5 Solution èø èø Term 2 3 We look at the initial value of the item and a ar ar ar ... calculate its new value each year. At the start, the 6 Where a = 10 and r = are the first term and the value of the item of jewellery is $1000 5 common ratio of a geometric progression. At the end of year 1, the value is $ 1000 + 10% ( $1000) To complete the job, 800 m3 of soil is to be removed. = $1000 ´ 1.1 Let n be the number of days taken to complete the job. Then S = 800 where S = sum of the first n At the end of year 2, the value is n n terms of the geometric progression. ($ 1000 ´ 1.1) + 10 % of ($ 1000 ´ 1.1) n = $ 1000 ´ (1.1)2 ar()-1 S = r >1 n r -1 At the end of year 3, the value is Substituting in the above formula, we have: ($ 1000 ´ 1.12) + 10 % of ($ 1000 ´ 1.12) n æöæö6 = $ 1000 ´ (1.1)3 10ç÷ç÷- 1 ç÷èø5 \èø=800 6 When we observe its new value each year, a -1 special pattern emerges. 5 n æöæö6 This is a geometric progression with first term of 10- 1= 160 ç÷ç÷ $ 1000 and common ratio of 1.1. èøèø5 1.2n = 17 Hence, the estimated value of the item of jewellery To solve for n, take logs to the base 10 after eleven years can be obtained by using the formula for the n th term of the GP. ��1.2 = ��17 ���1.2 = ��17 Tar= n - 1 ��17 n � = =$ 1000 ´ 1.111 – 1 = $ 1000 ´ 1.110 = $2 593.74 ��1.2 � = 15.5, � � � The value at the start of the 11th year is $2 593.74 Hence the job is complete on the 16th day

Example 12 www.faspassmaths.comSequences and series using Σ notation An earth mover has a job to remove 800 m3 of soil We can describe an AP by writing a formula for the over a certain period. On the first day 10 m3 is nth term of the sequence instead of listing the terms. moved and on each day the amount of soil moved Let Tn represent the nth term, where T = 2n −1 and n = 1,2,3,...(n is a positive integer) is times the amount removed on the previous n The terms of the A.P can be calculated by day. substituting n = 1, 2, 3, … to obtain: This continues until the job is complete. Find the number of days taken to complete the job. T1 =21() - 1 T2 =22() - 1 T3 =23() - 1

= 1 = 3 = 5

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 6 of 7 And so, the AP is {1, 3, 5, …} Solution The sum of this AP to infinity is a series and can be 40 written using the Σ notation as å()()()()()2r - 3= 2 1- 3+ 2 2- 3+ 2 3- 3++ ... 2 40- 3 r=1 n=∞ 40 ∑(2n −1) ()2r - 3=- 1++++ 1 3 ... 77 n=1 å r=1 n =∞ � = 1, � = 2, � = 40 and � = 77 So we can write (2n −1) = 1+ 3+ 5+ ... This is an AP with ∑ 40 n 1 = \-23rS= å() 40 r=1 Summation Notation b n n For an AP Saln =+() OR Sandn =+{21()-} A series may be denoted by åTn , where 2 2 na= 40 40 b (2r − 3) = (−1+ 77) = 20(76) = 1520 å ∑ 2 denotes the sum of and åTn denotes the sum r=1 na= OR of all the terms from n = a to n = b. 40 40 ∑(2r − 3) = {2(−1) + (40 −1)2} r=1 2 Example 13 = 20{−2 + (39 × 2)} = 20(76) = 1520 50 Find the sum å ()32k + . k = 1 Alternative Method We could have solved this problem using basic laws involving the summation notation. We can distribute Solution the summation as shown: 50 32k + means the sum of the terms of the å() 40 40 40 k = 1 ååå()232rr-º - 3 sequence from k = 1 to k = 50 , inclusive. rr==11r=1 To obtain the first term substitute k = 1 , so T 3 1 2 5 1 = () + = Expressing the first term as an AP, we have 40 T 3 2 2 8 r =+++1 2 3 ... + 40 The second and third terms are 2 = () + = , å r=1 and T = 3 3 + 2 = 11. a = 1 d = 1 n = 40 l = 40 3 () This is an AP with , , and . The sequence looks like {5, 8, 11, …} Any of the formulae for Sn may be used. 40 st This is an AP with 1 term a = 5 common S40 =+=()()1402041 2 difference, d =853-= and the number of terms, S = 820 n = 50 40 l =+=350 2 152 We now simplify the second term. The last term, () 40 40 50 n 50 å3=+++ 3 3 3 ... + 3 (40 times). So, å3= 120 ∴ 3k + 2 = S = (a + l) = 5+152 = 3925 r=1 r=1 ∑() 50 2 2 () k = 1 40 40 40 We could have used: ååå()232rr-º - 3 rr==11r=1 n www.faspassmaths.com40 S = 2a + n −1 d 50 2 { () } ()()2r - 3= 2 820- 120 å 50 r=1 = {2() 5 + ()50 −1 3} = 3925 40 2 å()2r - 3= 1520 r=1

Example 14 40 Evaluate å(2r - 3) . r =1

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