GEOMETRIC PROGRESSIONS

A Geometric Progression (GP) or Geometric Series is one in which each term is found by multiplying the previous term by a fixed number (common ratio). If the first term is denoted by a, and the common ratio by r, the series can be written as: a + e.g … Hence the n th term is given by: or 2 – –16 + … ar 2 + ar +ar 3 + …

The sum of the first n terms, S n is found as follows: S n = a + ar + ar 2 + ar 3 +…ar n–2 + ar n–1 …(1) Multiply throughout by r: r S n = ar + ar 2 + ar 3 + ar 4 + …ar n–1 + ar n …(2) Now subtract (2) – (1):r S n – S n = ar n – a Factorise:S n (r – 1) = a (r n – 1 ) Hence:

Example 1:For the series … Find a) The 10 th term. b) The sum of the first 8 terms. a) For the series, we have:a = 2, r = 3 Using: u n = ar n–1 u 10 = 2(3 9 )= = 6560

Example 2:For the series 32 – – … Find a) The 12 th term. b) The sum of the first 7 terms. a) For the series, we have:a = 32, r = Using: u n = ar n–1 u 12 = 32 = – ( ) 1 2 – 11 = 1 64 – We can write this as: (This ensures that the denominator is positive).

Example 3: a) Firstly, we need to find the first few terms: The series is:6We have a = 6, r = 2 The number of terms, n = 11 = b)= (3 1 – 20 ) + (3 2 – 20 ) + (3 3 – 20 ) + … = ( …) – ( …) A Geometric Series12 of these – (20 × 12)= – 240 = …

Example 4:In a Geometric Series, the third term is 36, and the sixth term is For the series, find the common ratio, the first term and the twentieth term. The third term is 36i.e. u 3 = 36 The sixth term is 121.5i.e. u 6 = Using: u n = ar n–1 ar 2 = 36 ….(1) ar 5 = ….(2) Now, divide equation (2) by equation (1): So r 3 = r = 1.5 Substitute this value into equation (1):a(1.5) 2 = 36a = 16 Now the 20 th term,u 20 = ar 19 = 16 (1.5) 19 = (To the nearest integer)

Example 5:The first three terms of a geometric progression are x, x + 3, 4x. Find the two possible values for the common ratio. For each value find these first three terms, and the common ratio. The ratio of a G.P. is found by dividing a term by the previous term: Now, x (4x) = (x + 3)(x + 3) 3x 2 – 6x – 9 = 0 Divide by 3:x 2 – 2x – 3 = 0 (x – 3)(x + 1) = 0 So, either x = 3,giving the terms: 3, 6, 12with ratio r = 2 or x = –1,giving the terms: –1, 2, –4with ratio r = –2 4x 2 = x 2 + 6x + 9

Example 6: £100 is invested into an account (earning 5% compound interest per annum), at the start of every year. Find the amount in the account at the end of the 8 th year. The amount in the account at the start of each year earns 5% interest. i.e. The amount is increased by 5%. To increase an amount by 5%,multiply by 1.05 At the end of the 1 st year The amount in the account: = (100 × 1.05) At the start of the 2 nd year = (100 × 1.05) At the end of the 2 nd year = [100 + (100 × 1.05)] × 1.05 = (100 × 1.05) + (100 × ) At the start of the 3 rd year = (100 × 1.05) + (100 × ) At the end of the 3 rd year = {100 + (100 × 1.05) + (100 × )} × 1.05 = (100 × 1.05) + (100 × ) + (100 × )

The amount we now have: At the end of the 3 rd year = (100×1.05) + (100× ) + (100× ) This is a GP With a = 100 × 1.05= 105 and r = 1.05 We want the sum to 8 terms,i.e. n = 8 = Hence, the amount in the account after 8 years is £ (To the nearest penny.)

A Geometric Series is one in which the terms are found by multiplying each term by a fixed number (common ratio). The n th term is given by: The sum of the first n terms is given by: In problems where r < 1 it is better to write the above as: Summary of key points: This PowerPoint produced by R. Collins; © ZigZag Education 2008–2010