MATHEMATIC TIME Arithmetic and geometricprogressions mcTY-apgp-2009-1 This unit introduces sequences and series, and gives some si mple examples of each. It also explores particular types of sequence known as arithmetic p rogressions (APs) and geometric progressions (GPs), and the corresponding series. In order to master the techniques explained here it is vital t hat you undertake plenty of practice exercises so that they become second nature. After reading this text, and/or viewing the video tutorial o n this topic, you should be able to: • recognise the difference between a sequence and a series; • recognise an arithmetic progression; • find the n -th term of an arithmetic progression; • find the sum of an arithmetic series; • recognise a geometric progression; • find the n -th term of a geometric progression; • find the sum of a geometric series; • find the sum to infinity of a geometric series with common ratio | r | < 1 . Contents 1. Sequences 2 2. Series 3 3. Arithmetic progressions 4 4. The sum of an arithmetic series 5 5. Geometric progressions 8 6. The sum of a geometric series 9 7. Convergence of geometric series 12 mathcentre.ac.uk 1 c math centre 2009 1. Sequences What is a sequence? It is a set of numbers which are written in s ome particular order. For example, take the numbers 1 , 3 , 5 , 7 , 9 , . . . . Here, we seem to have a rule. We have a sequence of odd numbers. To put this another way, we start with the number 1, which is an odd number, and then each s uccessive number is obtained by adding 2 to give the next odd number. Here is another sequence: 1 , 4 , 9 , 16 , 25 , . . . . This is the sequence of square numbers. And this sequence, 1 , − 1 , 1 , − 1 , 1 , − 1 , . . . , is a sequence of numbers alternating between 1 and − 1 . In each case, the dots written at the end indicate that we must consider the sequence as an infinite sequence, so that it goes on for ever. On the other hand, we can also have finite sequences. The numbe rs 1 , 3 , 5 , 9 form a finite sequence containing just four numbers. The numb ers 1 , 4 , 9 , 16 also form a finite sequence. And so do these, the numbers 1 , 2 , 3 , 4 , 5 , 6 , . . . , n . These are the numbers we use for counting, and we have include d n of them. Here, the dots indicate that we have not written all the numbers down explic itly. The n after the dots tells us that this is a finite sequence, and that the last number is n . Here is a sequence that you might recognise: 1 , 1 , 2 , 3 , 5 , 8 , . . . . This is an infinite sequence where each term (from the third te rm onwards) is obtained by adding together the two previous terms. This is called the Fibonacc i sequence. We often use an algebraic notation for sequences. We might ca ll the first term in a sequence u 1 , the second term u 2 , and so on. With this same notation, we would write u n to represent the n -th term in the sequence. So u 1 , u 2 , u 3 , . . . , u n would represent a finite sequence containing n terms. As another example, we could use this notation to represent the rule for the Fibonacci sequence. W e would write u n = u n − 1 + u n − 2 to say that each term was the sum of the two preceding terms. mathcentre.ac.uk 2 c math centre 2009 Key Point A sequence is a set of numbers written in a particular order. W e sometimes write u 1 for the first term of the sequence, u 2 for the second term, and so on. We write the n -th term as u n . Exercise 1 (a) A sequence is given by the formula u n = 3 n + 5 , for n = 1 , 2 , 3 , . . . . Write down the first five terms of this sequence. (b) A sequence is given by u n = 1 /n 2 , for n = 1 , 2 , 3 , . . . . Write down the first four terms of this sequence. What is the 10th term? (c) Write down the first eight terms of the Fibonacci sequence defined by u n = u n − 1 + u n − 2 , when u 1 = 1 , and u 2 = 1 . (d) Write down the first five terms of the sequence given by u n = ( − 1) n +1 /n . 2. Series A series is something we obtain from a sequence by adding all the terms together. For example, suppose we have the sequence u 1 , u 2 , u 3 , . . . , u n . The series we obtain from this is u 1 + u 2 + u 3 + . . . + u n , and we write S n for the sum of these n terms. So although the ideas of a ‘sequence’ and a ‘series’ are related, there is an important distinction bet ween them. For example, let us consider the sequence of numbers 1 , 2 , 3 , 4 , 5 , 6 , . . . , n . Then S 1 = 1 , as it is the sum of just the first term on its own. The sum of the fi rst two terms is S 2 = 1 + 2 = 3 . Continuing, we get S 3 = 1 + 2 + 3 = 6 , S 4 = 1 + 2 + 3 + 4 = 10 , and so on. mathcentre.ac.uk 3 c math centre 2009 Key Point A series is a sum of the terms in a sequence. If there are n terms in the sequence and we evaluate the sum then we often write S n for the result, so that S n = u 1 + u 2 + u 3 + . . . + u n . Exercise 2 Write down S 1 , S 2 , . . . , S n for the sequences (a) 1 , 3 , 5 , 7 , 9 , 11 ; (b) 4 , 2 , 0 , − 2 , − 4 . 3. Arithmetic progressions Consider these two common sequences 1 , 3 , 5 , 7 , . . . and 0 , 10 , 20 , 30 , 40 , . . . . It is easy to see how these sequences are formed. They each sta rt with a particular first term, and then to get successive terms we just add a fixed value to the pre vious term. In the first sequence we add 2 to get the next term, and in the second sequence we add 1 0. So the difference between consecutive terms in each sequence is a constant. We could al so subtract a constant instead, because that is just the same as adding a negative constant. F or example, in the sequence 8 , 5 , 2 , − 1 , − 4 , . . . the difference between consecutive terms is − 3 . Any sequence with this property is called an arithmetic progression , or AP for short. We can use algebraic notation to represent an arithmetic pro gression. We shall let a stand for the first term of the sequence, and let d stand for the common difference between successive terms. For example, our first sequence could be written as 1, 3, 5, 7, 9, ... 1, 1 + 2 , 1 + 2 × 2 , 1 + 3 × 2 , 1 + 4 × 2 , ... , and this can be written as a, a + d, a + 2 d, a + 3 d, a + 4 d, . . . where a = 1 is the first term, and d = 2 is the common difference. If we wanted to write down the n -th term, we would have a + ( n − 1) d , mathcentre.ac.uk 4 c math centre 2009 because if there are n terms in the sequence there must be ( n − 1) common differences between successive terms, so that we must add on ( n − 1) d to the starting value a . We also sometimes write ℓ for the last term of a finite sequence, and so in this case we wou ld have ℓ = a + ( n − 1) d . Key Point An arithmetic progression, or AP, is a sequence where each ne w term after the first is obtained by adding a constant d , called the common difference , to the preceding term. If the first term of the sequence is a then the arithmetic progression is a, a + d, a + 2 d, a + 3 d, . . . where the n -th term is a + ( n − 1) d . Exercise 3 (a) Write down the first five terms of the AP with first term 8 and c ommon difference 7. (b) Write down the first five terms of the AP with first term 2 and c ommon difference − 5 . (c) What is the common difference of the AP 11 , − 1 , − 13 , − 25 , . . . ? (d) Find the 17th term of the arithmetic progression with firs t term 5 and common differ- ence 2. (e) Write down the 10th and 19th terms of the APs (i) 8 , 11 , 14 , . . . , (ii) 8 , 5 , 2 . . . . (f) An AP is given by k, 2 k/ 3 , k/ 3 , 0 , . . . . (i) Find the sixth term. (ii) Find the n th term. (iii) If the 20th term is equal to 15, find k . 4. The sum of an arithmetic series Sometimes we want to add the terms of a sequence. What would we get if we wanted to add the first n terms of an arithmetic progression? We would get S n = a + ( a + d ) + ( a + 2 d ) + . . . + ( ℓ − 2 d ) + ( ℓ − d ) + ℓ . Now this is now a series, as we have added together the n terms of a sequence. This is an arithmetic series , and we can find its sum by using a trick. Let us write the series down again, but this time we shall write it down with the terms in reverse o rder. We get S n = ℓ + ( ℓ − d ) + ( ℓ − 2 d ) + . . . + ( a + 2 d ) + ( a + d ) + a . mathcentre.ac.uk 5 c math centre 2009 We are now going to add these two series together. On the left- hand side, we just get 2 S n . But on the right-hand side, we are going to add the terms in the two series so that each term in the first series will be added to the term vertically below it in th e second series. We get 2 S n = ( a + ℓ ) + ( a + ℓ ) + ( a + ℓ ) + . . . + ( a + ℓ ) + ( a + ℓ ) + ( a + ℓ ) , and on the right-hand side there are n copies of ( a + ℓ ) so we get 2 S n = n ( a + ℓ ) . But of course we want S n rather than 2 S n , and so we divide by 2 to get S n = 1 2 n ( a + ℓ ) . We have found the sum of an arithmetic progression in terms of its first and last terms, a and ℓ , and the number of terms n . We can also find an expression for the sum in terms of the a , n and the common difference d . To do this, we just substitute our formula for ℓ into our formula for S n . From ℓ = a + ( n − 1) d , S n = 1 2 n ( a + ℓ ) we obtain S n = 1 2 n ( a + a + ( n − 1) d ) = 1 2 n (2 a + ( n − 1) d ) . Key Point The sum of the terms of an arithmetic progression gives an ari thmetic series. If the starting value is a and the common difference is d then the sum of the first n terms is S n = 1 2 n (2 a + ( n − 1) d ) . If we know the value of the last term ℓ instead of the common difference d then we can write the sum as S n = 1 2 n ( a + ℓ ) . Example Find the sum of the first 50 terms of the sequence 1 , 3 , 5 , 7 , 9 , . . . . mathcentre.ac.uk 6 c math centre 2009 Solution This is an arithmetic progression, and we can write down a = 1 , d = 2 , n = 50 . We now use the formula, so that S n = 1 2 n (2 a + ( n − 1) d ) S 50 = 1 2 × 50 × (2 × 1 + (50 − 1) × 2) = 25 × (2 + 49 × 2) = 25 × (2 + 98) = 2500 . Example Find the sum of the series 1 + 3 5 + 6 + 8 5 + . . . + 101 . Solution This is an arithmetic series, because the difference between the terms is a constant value, 2 5 . We also know that the first term is 1, and the last term is 101. Bu t we do not know how many terms are in the series. So we will need to use the formula for t he last term of an arithmetic progression, ℓ = a + ( n − 1) d to give us 101 = 1 + ( n − 1) × 2 5 . Now this is just an equation for n , the number of terms in the series, and we can solve it. If we subtract 1 from each side we get 100 = ( n − 1) × 2 5 and then dividing both sides by 2 5 gives us 40 = n − 1 so that n = 41 . Now we can use the formula for the sum of an arithmetic progre ssion, in the version using ℓ , to give us S n = 1 2 n ( a + ℓ ) S 41 = 1 2 × 41 × (1 + 101) = 1 2 × 41 × 102 = 41 × 51 = 2091 . mathcentre.ac.uk 7 c math centre 2009 Example An arithmetic progression has 3 as its first term. Also, the su m of the first 8 terms is twice the sum of the first 5 terms. Find the common difference. Solution We are given that a = 3 . We are also given some information about the sums S 8 and S 5 , and we want to find the common difference. So we shall use the formul a S n = 1 2 n (2 a + ( n − 1) d ) for the sum of the first n terms. This tells us that S 8 = 1 2 × 8 × (6 + 7 d ) . and that S 5 = 1 2 × 5 × (6 + 4 d ) So, using the given fact that S 8 = 2 S 5 , we see that 1 2 × 8 × (6 + 7 d ) = 2 × 1 2 × 5 × (6 + 4 d ) 4 × (6 + 7 d ) = 5 × (6 + 4 d ) 24 + 28 d = 30 + 20 d 8 d = 6 d = 3 4 . Exercise 4 (a) Find the sum of the first 23 terms of the AP 4 , − 3 , − 10 , . . . . (b) An arithmetic series has first term 4 and common difference 1 2 . Find (i) the sum of the first 20 terms, (ii) the sum of the first 100 terms. (c) Find the sum of the arithmetic series with first term 1, com mon difference 3, and last term 100. (d) The sum of the first 20 terms of an arithmetic series is iden tical to the sum of the first 22 terms. If the common difference is − 2 , find the first term. 5. Geometric progressions We shall now move on to the other type of sequence we want to exp lore. Consider the sequence 2 , 6 , 18 , 54 , . . . . Here, each term in the sequence is 3 times the previous term. A nd in the sequence 1 , − 2 , 4 , − 8 , . . . , each term is − 2 times the previous term. Sequences such as these are called geometric progres- sions , or GPs for short. mathcentre.ac.uk 8 c math centre 2009 Let us write down a general geometric progression, using alg ebra. We shall take a to be the first term, as we did with arithmetic progressions. But here, ther e is no common difference. Instead there is a common ratio, as the ratio of successive terms is al ways constant. So we shall let r be this common ratio. With this notation, the general geometri c progression can be expressed as a, ar, ar 2 , ar 3 , . . . . So the n -th can be calculated quite easily. It is ar n − 1 , where the power ( n − 1) is always one less than the position n of the term in the sequence. In our first example, we had a = 2 and r = 3 , so we could write the first sequence as 2 , 2 × 3 , 2 × 3 2 , 2 × 3 3 , . . . . In our second example, a = 1 and r = − 2 , so that we could write it as 1 , 1 × ( − 2) , 1 × ( − 2) 2 , 1 × ( − 2) 3 , . . . . Key Point A geometric progression, or GP, is a sequence where each new t erm after the first is obtained by multiplying the preceding term by a constant r , called the common ratio . If the first term of the sequence is a then the geometric progression is a, ar, ar 2 , ar 3 , . . . where the n -th term is ar n − 1 . Exercise 5 (a) Write down the first five terms of the geometric progressio n which has first term 1 and common ratio 1 2 . (b) Find the 10th and 20th terms of the GP with first term 3 and co mmon ratio 2. (c) Find the 7th term of the GP 2 , − 6 , 18 , . . . , 6. The sum of a geometric series Suppose that we want to find the sum of the first n terms of a geometric progression. What we get is S n = a + ar + ar 2 + ar 3 + . . . + ar n − 1 , and this is called a geometric series . Now the trick here to find the sum is to multiply by r and then subtract: S n = a + ar + ar 2 + ar 3 + . . . + ar n − 1 rSn = ar + ar 2 + ar 3 + . . . + ar n − 1 + ar n S n − rS n = a − ar n mathcentre.ac.uk 9 c math centre 2009 so that S n (1 − r ) = a (1 − r n ) . Now divide by 1 − r (as long as r 6 = 1 ) to give S n = a (1 − r n ) 1 − r . Key Point The sum of the terms of a geometric progression gives a geomet ric series. If the starting value is a and the common ratio is r then the sum of the first n terms is S n = a (1 − r n ) 1 − r provided that r 6 = 1 . Example Find the sum of the geometric series 2 + 6 + 18 + 54 + . . . where there are 6 terms in the series. Solution For this series, we have a = 2 , r = 3 and n = 6 . So S n = a (1 − r n ) 1 − r S 6 = 2(1 − 3 6 ) 1 − 3 = 2(1 − 729) − 2 = − ( − 728) = 728 . Example Find the sum of the geometric series 8 − 4 + 2 − 1 + . . . where there are 5 terms in the series. mathcentre.ac.uk 10 c math centre 2009 Solution For this series, we have a = 8 , r = − 1 2 and n = 5 . So S n = a (1 − r n ) 1 − r S 5 = 8(1 − − 1 2 5 ) 1 − − 1 2 = 8(1 − − 1 32 ) 3 2 = 2 × 8 × 33 32 3 = 11 2 = 5 1 2 . Example How many terms are there in the geometric progression 2 , 4 , 8 , . . . , 128 ? Solution In this sequence a = 2 and r = 2 . We also know that the n -th term is 128. But the formula for the n -th term is ar n − 1 . So 128 = 2 × 2 n − 1 64 = 2 n − 1 2 6 = 2 n − 1 6 = n − 1 n = 7 . So there are 7 terms in this geometric progression. Example How many terms in the geometric progression 1 , 1 1 , 1 21 , 1 331 , . . . will be needed so that the sum of the first n terms is greater than 20? Solution The sequence is a geometric progression with a = 1 and r = 1 1 . We want to find the smallest value of n such that S n > 20 . Now S n = a (1 − r n ) 1 − r , mathcentre.ac.uk 11 c math centre 2009
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