Geometric Sequences and Series

Sequences are called geometric sequences if and only if we can get from one term to the next, by multiplying the same number each time. This number is called the common ratio, r. e.g. 1, 2, 4, 8, 16, … r = 2 100, 25, 6.25, 1.5625, … r = ¼ 2, -6, 18, -54, 162, … r = -3

Each individual number in the sequence is called U. Therefore the sequence consists of

U1, U2, U3, U4 … etc.

To find any common ratio, just divide any U by its predecessor. 푈2 푈3 푈9 e.g. or or 푈1 푈2 푈8

A common ratio can be negative, e.g. 2, -6, 18, -54, 162, … r = -3 Or a fraction, e.g. 100, 25, 6.25, 1.5625, … r = ¼ Or both, 1 1 e.g. 90, -30, 10, -3 , … r = - 3 3

Important: Sometimes a geometric sequence is called a geometric progression.

In a geometric sequence the first term is called a and the common ratio is called r. a, ar1, ar2, ar3, … arn-1 1st 2nd 3rd 4th … nth term

The common ratio r is one less than its position in the sequence. So e.g. the 4th term of the geometric sequence is ar3 and, The nth term is arn-1.

E1): find the i 10th and ii nth terms.

a) 3, 6, 12, 24, … b) 40, -20, 10, -5, …

A1): a) i 10th term = 3 × (2)9 as r = 6/3 = 2 = 3 × 512 = 1536 ii nth term = 3 × 2n-1 b) 1 i 10th term = 40 × (- )9 as r = -20/40 = -½ 2 1 = 40 × (- ) 512 5 = 64 ii nth term = 40 × (-½)n-1 = 5 × 8 × (-½)n-1 = 5 × 23 × (-½)n-1 n-1 5 n-4 = (-1) × /2

E2): The second term of a geometric sequence is 4 and the 4th term is 8. Find the exact values of a) the common ratio, b) first term and c) the 10th term. A2):

a) As the 2nd term is 4 ar = 4

And the 4th term is ar3 = 8

arr2= 8

ar = 4 ∴ 4r2= 8

r2 = 2

r = 2

b) ar = 4

a 2 = 4

4 a = 2

4 2 a = 2 2

4 2 a = 2

a = 2 2

c) 10th term = ar9

ar9 = 2 2 ( 2)9

ar9 = 2 ( 2)10

ar9 = 2 (2)5

ar9 = 26 = 64

E3): 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?

2 3 4 A3): a ar ar ar ar What we get from the question is: ar4 = 10,000 r = 4% = 1.04 a = £A ∴ ar = £A × 1.04 ar2 = £A × 1.042 Remember after 5 years (it means the 5th term) investment is worth: ∴ ar5 = £A × 1.045 A × 1.045 = 10,000 10.000 A = 1.045 A = £8219.27 So initial investment = £8219.27 After 10 years ar10 = £8219.27 × 1.0410 10.000 Or = ( ) × 1.0410 = 10 000 × 1.045 1.045 = £12166.53

E4): What is the first term in the geometric progression 3, 6, 12, 24, … to exceed 1 million?

A4): nth term = arn-1 = 3 × (2)n-1 We want nth term > 1,000,000 So 3 × 2n-1 > 1,000,000 1,000,000 2n-1 > taking log of both sides 3 1,000,000 (n-1) × log(2) > log( ) 3 1,000,000 푙표𝑔 ( ) 5.52287875 (n-1) > 3 = = 18.35 log ⁡(2) 0.30102999

(n-1) > 18.35 (We calculated formula by assuming = instead of >)

(And as n has to be an integer) ∴ n - 1 = 19 (We put the formula equal to 1,000,000 then answer of n-1 = 19) (Now the question says: What is the first term to exceed 1 million?)

As equal to 1,000,000 n-1 = nth term = 19

Then to exceed 1 million n-1 = nth term = 20 ∴ the 20th term is the first to exceed 1,000,000.

E5): A population of ants is growing at a rate of 10% a year. If there were 200 ants in the initial population, write down the number after,

a) 1 year b) 2 years c) 3 years d) 10 years

A5): r = 1.10 a = 200 a) ar = 220 2 b) ar = 242 3 c) ar = 266 10 d) ar = 519

E6): A motorcycle has four gears. The maximum speed in bottom gear is 40 Km h-1 and the maximum speed in top gear is 120 Km h-1. Given that the maximum speed in each successive gear form a geometric progression, calculate, in Km h-1 to one decimal place, the maximum speeds in the two intermediate gears.

A6): a ar ar2 ar3 40 ? ? 120

a = 40 ar3 = 120 ∴ 40r3 = 120 120 r3 = = 3 40 r = 1.44

ar = 40 × 1.44 ar = 57.8

ar2 = 82.94

E7): A car depreciates in value by 15% a year. If it is worth £11,054.25 after 3 years, what was its new price and when will it first be worth less than £5,000?

A7): r = 1 - 0.15 depreciates U3 = £11054.25 a = ? a ar ar2 ar3 £? 11054.25 r = 1 - 0.15 = 0.85 ar3 = 11054.25 a(0.85)3 = 11054.25 a × 0.614125 = 11054.25 a = £18,000

arn-1 < 5,000 18,000 (0.85)n-1 < 5,000 5,000 (0.85)n-1 < 18,000 5,000 Log (0.85)n-1 < log ( ) 18,000 (n-1) log (0.85) < log (0.2777778) log (0.2777778 ) (n-1) < log (0.85) −0.5563025 (n-1) < −0.0705811 n – 1 < 7.88 (We calculated formula by assuming = instead of <) (We put the formula equal to 5,000 then answer of n-1 = 7.88) (Now the question says: when will it first be worth less than £5,000?) As time depreciates the price of the car ∴ we have to use the word after 7.88 years.

E8): The population decline in a school of whales can be modelled by a geometric progression. Initially they were 80 whales in the school. Four years later there were 40. Find out how many there will be at the end of the fifth year. (Round to the nearest number)

A8): a = 80 ar4 = 40 5 ar = ? (End of year5 or beginning of year5 (the same)) 80(r)4 = 40 40 1 r4 = = 80 2 r = 0.84089 ar5 = 80 (0.84089)5 ar5 = 33.64 ≅ 34

E9): Fined which term in the progression 3, 12, 48, … is the first to exceed 1,000,000

A9): a = 3 r = 4 arn-1 > 1,000,000 3(4)n-1 > 1,000,000 1,000,000 4n-1 > 3 1,000,000 (n-1) log4 > log 3 1,000,000 log n-1 > 3 log 4 5.5228787 n-1 > 0.6020599 n-1 > 9.17 (And as n has to be an integer) ∴ n - 1 = 10 AS before to exceed means 1 more. ∴ 11th term is the first to exceed 1,000,000

E10): A virus is spreading such that the number of people infected increases by 4% a day. Initially 100 people were diagnosed with the virus. How many days will it be before 1,000 are infected?

A10): a = 100 r = 1.04 arn-1 = 1,000 100 Х 1.04n-1 = 1,000 1.04n-1 = 10 (n-1) log (1.04) = log(10) log ⁡(10) 1 (n-1) = = log ⁡(1.04) 0.0170333 (n-1) = 58.71 (And as n has to be an integer) (n-1) = 59 (In this formula we have not used any < or > sign) ∴ the answer 59 is the exact answer to the question.

E11): I invest £A in the bank at a rate of interest of 3.5% per annum. How long will it be before I double my money?

A11): a = £A r = 1.035 arn-1 = 2a Arn-1 = 2A 1.035n-1 = 2 (n-1) log(1.035) = log(2)

log ⁡(2) 0.30103 (n-1) = = log ⁡(1.035) 0.0149403 (n-1) = 20.149 n = 20.149 + 1 n = 21.149

E12): The fish in a particular area of the North Sea are being reduced by 6% each year due to over fishing. How long would it be before the fish stocks are halved?

A12):

a = F r = 1 – 0.06 = 0.94 (reduced by 6%) 푎 arn-1 = 2 퐹 Frn-1 = 2 1 0.94n-1 = 2 1 (n-1) log(0.94) = log( ) 2 1 log ⁡( ) −0.30103 (n-1) = 2 = log ⁡(0.94) −0.0268721 (n-1) = 11.2 11.2 years before the stock is halved.

The sum of a geometric series

If Sn = a + ar + ar2 + ar3 + … arn-2 + arn-1 r Sn = ar + ar2 + ar3 + ar4 + … arn-1 + arn Sn – r Sn = a – arn S (1 – r) = a (1 – rn)

풂 (ퟏ – 풓풏) S = ퟏ−풓 So the general formula for sum is:

풂 (ퟏ – 풓풏) S = um ퟏ−풓 Or 풂 (풓풏− ퟏ) S = um 풓−ퟏ

E13): Find the sum of the following series:

1024, -512, +256, -128, …. , +1.

A13): −512 1 a = 1042 r = = - 1024 2 arn-1 = +1 1 1024 (- )n-1 = 1 2 1 1 (- )n-1 = 2 1024 Or (-2)n-1 = 1024 (n-1) log (-2) = log 1024 log 1024 n - 1 = log (−2) We don’t have log of –Ve so -2 must be +2

log 1024 n - 1 = log (+2) 3.0103 n - 1 = 0.30103 n - 1 = 10 n = 1 + 10 n = 11

a(1 – rn ) As S = n 1 – r 1 1024 (1 – (− )11 ) 2 1024 (0.9995) Sn = 1 = 1 – (− ) 1.5 2

Sn = 683

The sum of a geometric series when having an investment in a regular and equal manners

In a geometric series sometimes you have an accumulation of an investment in regular and equal manners. For example if you invest some money (e.g. £2,000) each year in your bank and at the end of each year you get an interest of so many percent (e.g. 4%) then: To find the some of this geometric series 푟 (푟푛 −1) you have s = a n 푟−1 I could give you another example such that if you invest in your cellar some equal amount of wine each year (e.g. a barrel of

160 litres a year) while consuming some each year (e.g. 15% each year) then: To find the some of this geometric series you have 푟 (푟푛 − 1) s = a n 푟 − 1 The answer to the above example will be: 0.85 (0.8520 − 1) S = 160 20 0.85 − 1 = 160 (5.447) = 871.52 = 872 (close to nearest litre) Remember these types of questions could be a bit more complicated: Example1: What if your consumption of wine was 15 litres rather than 15%? Answer1: You should calculate what % of your consumption is equal to 15 litres (i.e. ((15 ÷ a) x 100) 15 That is: x 100 = 9.375 ≈ 9.38% 160 (i.e. the rate of 9.38% means r = 1 – 0.0938 r = 0.9062) 20

E14): An investor invests £2,000 on January 1st every year in a saving account that guaranties him 4% per annum for life. If interest is calculated on the 31st of December each year, how much will be in the account at the end of 10th year?

A14): Start of year 1, amount = (0) + 2,000 End of year 1, amount = (2,000) X 1.04 Start of year 2, amount = (2,000 X 1.04) + 2,000 End of year 2, amount = (2,000 X 1.04 + 2,000) X 1.04 = 2,000 X 1.042 + 2,000 X 1.04 Start of year 3, amount = (2,000 X 1.042 + 2,000 X 1.04) + 2,000 End of year 2, amount = (2,000 X 1.042 + 2,000 X 1.04 + 2,000) X 1.04 = 2,000 X 1.043 + 2,000 X 1.042 + 2,000 X 1.04 And so on, therefore by for the end of year 10: End of year 10 amount = 2,000 X 1.0410 + 2,000 X 1.049 + … + 2,000 X 1.04 Factorising 2,000 amount = 2,000 (1.0410 + 1.049 + … + 1.04) Now you see that inside bracket is a geometric series. Where a = 1.04 r = 1.04 풂 (풓풏− ퟏ) ∴Sn = 2,000 ( ) 풓−ퟏ 1.04 (1.0410 −1) S = 2,000 X 10 1.04−1

S10 = 2,000 X 12.486 = £24972.70

E15): Find the least value of n such that the sum of 1 + 2 + 4 + 8 + … to n terms would exceed 2,000,000.

1 (2n − 1) 1 (2n − 1) Sum to n terms is S = = A15): n 2−1 1 ∴ = 2n – 1 If this is to exceed 2,000,000 then Sn > 2,000,000 2n – 1 > 2,000,000 2n > 2,000,001 n log 2 > log 2,000,001

푙표𝑔 2,000,001 n > log 2 n > 20.9 It needs 21 terms to exceed 2,000,000

We can use this sign ∑ to show the sum of 푦 anything from x to y (i.e. ∑푥(푎푛푦푡ℎ𝑖푛𝑔)) 10 r e.g. S10 = ∑푟=1(3 × 2 ) = 3 × 21 + 3 × 22 + 3 × 23 + … + 3 × 210 = 3 (21 + 22 + 23 + … + 210)

210 − 1 = 3 × 2 ( ) 2 − 1

S10 = 6133

The sum to infinity of a convergent geometric series

In a series of s = 3 + 1.5 + 0.75 + 0.375 + … As n gets larger, s becomes closer to 6. So: we say that this infinite series is convergent, and has a sum to infinity of 6. (Convergent means that the series tends towards a specific value as more terms are added) When the terms of sequence in a series is getting smaller such as example of s then it is a convergent series. This happens because -1 < r < 1

So we can say The sum to infinity of a series exists only if: -1 < r < 1

The sum to infinity = S∞ a (1 − rn ) S = n 1−r If -1 < r < 1 then rn → 0 as n →∞ a (1 − 0) a So = 1 − r 1 − r ∴ The sum to infinity of a geometric series is: a if -1 < r < 1 i.e. r < 1 1−r

E16): The sum to 4 terms of a geometric series is 15 and the sum to infinity is 16.

a) Find the possible values of r. b) Given that the terms are all positive, find the first term in the series.

A16):

1− r4 a( ) = 15 (equation 1) a) 1−r a = 16 (equation 2) 1−r From equation 2 a = 16 (1 – r) 1− r4 Substitute in equation 1: 16 (1 – r) = 15 1−r 16 (1 – r4) = 15 15 1 – r4 = 16 1 r4 = 16 1 r = ± 2

1 b) As all the terms are positive, r = + 2 1 Substitute r = + back into (equation 2) 2 a = 16 1−r 1 16 (1 – ) = a 2 a = 8 ∴ The first term in the series is 8