Sequences Revision Learning Objective: Arithmetic Sequences Geometric Sequences Nth terms Sums
Arithmetic Sequences An arithmetic sequence has an recurrence relationship of the form: u 1 = a number u n+1 = u n + d The number added on each time is called the common difference.
Arithmetic Sequences Which of these are arithmetic sequences? u n+1 = u n - 4 u n+1 = u n + 11 u n+1 = 11u n - 4 u n+1 = 2u n - 4 u n+1 = (u n ) 2
Geometric Sequences A geometric sequence has an inductive definition of the form: u 1 = au n+1 = r u n The number multiplied by each time is called the common ratio.
Geometric Sequences The sequence 3, 12, 48, 192, 768 can be defined by… u 1 =3u n+1 =4 u n u 2 = 3 x 4u 2 = 12 u 3 = 3 x 4 x 4u 3 = 48 u 4 = 3 x 4 x 4 x 4u 4 = 192 u n =3 x 4 n-1 1 st termCommon ratio
Sequences For the sequence : 6, 11, 16, ……, 731 How many terms? u n = u 1 + (n-1)d 731 =6 + 5(n-1) u 1 =d =56u n =731 n -1 =(731-6)/5 = 145 n = 146 What sort of question? nth term of arithmetric sequence
Sequences If I put £400 in a bank account on my 16 th birthday and get 5% interest per year. How much money would I have on my 41 st birthday? u n = u 1 x r n-1 u n =400 x u 1 =r = n =25 u n =£ What sort of question? nth term of geometric sequence
Sequences For the sequence: 7, 9, 11, 13, ….. What is the sum of the first 50 terms? What sort of question? Sum of terms in arithmetic sequence S n = n/2 (2u 1 + (n-1)d) S 50 = 25( x 2) u 1 =d =27n = S 50 = 25 x 112 =
Sequences The sequence: 2, 4a, 8a 2, …. Find an expression for the sum of the first 10 terms S 10 = u 1 (r n – 1)/(r-1) S 10 =2((2a) 10 -1)/(2a-1) u 1 =r =2a2n =10 S 10 =(2 11 a 10 -2)/(2a-1) What sort of question? sum terms of geometric sequence
The 1 st term of a geometric sequence is 12 and the sum to Infinity is 9. Find the ratio of terms u 1 = = 9(1-r) 12 = 9 - 9r 9r = = -3 r = - 1 / 3
Converging series A geometric series converges to a limit when.. Means the size of r including negatives
Converging series Which of the following values of r will converge to a limit?
Example 3 r = Evaluate: 1/21/2 1 st term, when n=0 u 1 = ( 1 / 2 ) 0 = 1
Different sorts of question to try As a fraction Sum to infinity / / 25 + …r = u 1 = 2/52/5 1/31/3 1 / 10 1 (1/3)1=1/3(1/3)1=1/3 0.8= 4 / 5 sum = 1 / 3 / 5 = 5 / 3 sum = 1 / 3 / 2 / 3 = 1 / 2 sum = 4 / 5 / 9 / 10 = 8 / 9