Geometric Sequence r=5 Eg 2, 10, 50, =5 2 50= =5 50

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Presentation transcript:

Geometric Sequence r=5 Eg 2, 10, 50, 250 ... 10=5 2 50=5 10 250=5 50 A geometric sequence is one where to get from one term to the next you multiply by the same number each time. This number is called the common ratio, r. 1 2 3 4 Eg 2, 10, 50, 250 ... r=5 x5 x5 x5 10=5 2 50=5 10 250=5 50

‘Second term divided by the first’ Geometric Sequence 1 2 3 4 Eg 90, -30, 10, ... r= x x x 10 = -30 = 10 -30= 90 Common ratio =u2 u1 ‘Second term divided by the first’

How do we find the nth term? r= the number you times by to get to the next term a= the first term of the sequence 1 2 3 4 n Eg 2, 10, 50, 250 ... ? ... r= 5 a=2 a ar ar2 ar3 arn-1 This is the same for all geometric sequences

Your check list You will need to find or use these: r= the number you times by to get to the next term a= the first term of the sequence What terms have you got/need to find? The second term of a geometric sequence is 4 and the 4th term is 8. Find the values of a) the common ration b) first term and c) the 10th term. So what do we have: 1 2 3 4 ?, 4, ?, 8 .... ar ar3

?, 4, ?, 8 .... 1) ar =4 2) ar3 =8 ar ar3 ar3 =8 ar 4 r2 =2 r =√2 The second term of a geometric sequence is 4 and the 4th term is 8. Find the values of a) the common ration b) first term and c) the 10th term. r= the number you times by to get to the next term a= the first term of the sequence What terms have you got/need to find? So what do we have: 1 2 3 4 ?, 4, ?, 8 .... 1) ar =4 2) ar3 =8 ar ar3 ar3 =8 ar 4 r2 =2 r =√2 Using ar= 4 a√2 = 4 a= 4 √2 a= 2 √2

ar9= 2 √2 (√2) 9 ar9= 2 √2 10 ar9= 2 x2 5 ar9= 2 6 10th term= 64 r= the number you times by to get to the next term a= 2 √2 a= the first term of the sequence What terms have you got/need to find? the 10th term nth term = arn-1 1 2 3 4 ...... 10 ?, 4, ?, 8 .... ar9 ar9= 2 √2 (√2) 9 ar9= 2 √2 10 ar9= 2 x2 5 ar9= 2 6 10th term= 64

r=1.04 ( think if it will increase in value you need that 1!) Andy invests £A at a rate of interest 4% per annum. After 5 years it will be worth £10 000. How much (to the nearest penny) will it be worth after 10 years? r= the number you times by to get to the next term a= the first term of the sequence What terms have you got/need to find? So what do we have: r=1.04 ( think if it will increase in value you need that 1!) a= £A (this gives you a hint that you will need to work out A) Think about the terms: 1 2 3 4 a ? ? ? Think about the terms: 1 2 3 4 5 6 a ? ? ? ? 10000 Think about the terms: 1 2 3 4 5 a ? ? ? ? Think about the terms: 1 2 3 a ? ? Think about the terms: 1 2 a ? Think about the terms: 1 a 4 year 1 year 2 year 3 year 5 year ar5=10 000 ( Power is usually the same as the years but always check!) ar10=? ( Using the same idea after 10 years will be the 11th term)

ax1.045=10 000 ( Replace the r!) ax1.045=10 000 ( Replace the r!) Andy invests £A at a rate of interest 4% per annum. After 5 years it will be worth £10 000. How much (to the nearest penny) will it be worth after 10 years? r= the number you times by to get to the next term a= the first term of the sequence What terms have you got/need to find? ax1.045=10 000 ( Replace the r!) ax1.045=10 000 ( Replace the r!) £A= £10 000 £A = £8219.27 1.045 ar10=? £10 000 x 1.0410 = £10 000 x 1.045 1.045 ar10= £12166.53

r=2 ( 6 divided by 3) a= 3 nth term = arn-1 > 1000 000 What is the first term in the geometric progression 3,6,12,24.... To exceed 1 million? r= the number you times by to get to the next term a= the first term of the sequence What terms have you got/need to find? r=2 ( 6 divided by 3) a= 3 nth term = arn-1 > 1000 000 3x2n-1 > 1000 000 2n-1 > 1000 000 3 n-1> log2 (1000 000) 3 n-1> 18.35 (2dp) n> 19.35 (2dp) n = 20 So 20th term

A geometric series is the sum of a geometric sequence Terms: 1 2 3 4 n-1 n Sn a + ar + ar2 + ar3 +.......+ arn-2 + arn-1 Terms: 1 2 3 4 n-1 n rSn ar + ar2 + ar3 + ar4 +.......+ arn-1 + arn Sn a + ar + ar2 + ar3 +.......+ arn-2 + arn-1 rSn ar + ar2 + ar3 + ar4 +.......+ arn-1 + arn Sn- r Sn = a-arn Sn(1-r)= a(1-rn) Sn = a(1-rn) 1-r

Geometric Series Sn = a(1-rn) 1-r Sn = a(rn-1) r-1 An investor invests £2000 on January 1st every year in an account that guarantees 4% per annum, If the interest is calculated on the 31st of December each year, how much will be in the account at the end of the 10th year? So using logic break it down to understand what is happening: End of year 1: 2000 x 1.04 Start of year 2: 2000 x 1.04 +2000 End of year 2: (2000 x 1.04 +2000) x 1.04 2000 x 1.042 +2000 x 1.04 Start of year 3: 2000 x 1.042 +2000 x 1.04 +2000 Start of year 3: (2000 x 1.042 +2000 x 1.04 +2000) x 1.04 2000 x 1.043 +2000 x 1.042 +2000 x 1.04

Geometric Series Sn = a(rn-1) r-1 Geometric series!!! r=1.04 a=1.04 End of year 1: 2000 x 1.04 Sn = a(rn-1) r-1 Start of year 2: 2000 x 1.04 +2000 End of year 2: (2000 x 1.04 +2000) x 1.04 2000 x 1.042 +2000 x 1.04 Start of year 3: 2000 x 1.042 +2000 x 1.04 +2000 Start of year 3: (2000 x 1.042 +2000 x 1.04 +2000) x 1.04 2000 x 1.043 +2000 x 1.042 +2000 x 1.04 End of year 10: 2000 x 1.0410 +2000 x 1.049 +..............+2000 x 1.04 End of year 10: 2000 (1.0410 + 1.049 +..............+1.04) Geometric series!!! r=1.04 a=1.04 End of year 10: 2000 x 1.04 (1.0410-1) 1.04-1 End of year 10: = £24 972.70

Geometric Series Sn = a(rn-1) r-1 Some problems could be described like this The sum of Sn = a(rn-1) r-1 This for... ..... r=1 to 10 = 3x21 +3x22 +3x23 +3x24 +3x25 +3x26 +3x27 +3x28 +3x29 +3x210 =3(21 +22 +23 +24 +25 +26 +27 +28 +29 +210) Geometric series!!! r=2 a=2 = 3 x 2 (210 -1) 2-1 S10 = 6138

Sum to infinity of a convergent Geometric series Convergent means that the series tends towards a specific value as more terms are added. This value is called the limit. Consider this series S= 3 + 1.5 + 0.75 + 0.375 + ... r= 1 2 a=3 Test for different values of n: As n gets larger S becomes closer to 6

Sum to infinity of a convergent Geometric series This series S= 3 + 1.5 + 0.75 + 0.375 + ... Is a convergent series. This happens because: -1< r >1 If this were not true it would not be convergent. Sn = a(1-rn) 1-r But if -1< r >1, as S∞ = a(1-0) 1-r S∞ = __a__ 1-r

nth term = arn-1 Sn = a(1-rn) 1-r S∞ = __a__ 1-r