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Sequences & Series

All the maths stuff
by

Quintin Walker

on 2 December 2012

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Transcript of Sequences & Series

Sequences & Series Learning Outcomes Recap on the grade 11 curriculum. RECAP !!! LINEAR QUADRATIC EXPONENTIAL Types of Number Patterns Number Patterns arrangement of numbers according to some RULE responsible to generate the numbers in the pattern 3 ; 5 ; 7 ; 9 ; 11 ; 13 2 2 2 2 2 aka - arithmetic
sequences T = an + b n 6 ; 9 ; 12 ; 15 ; 18 3 3 3 3 common difference term first difference unknown 1 term 2 term 6 ; 9 ; 12 ; 15 ; 18 General term is T = an + b n First Difference is 3 > a = 3 T = 3n + b n for n = 3 therefore T = 12 3 T = 3 (3) + b 3 12 = 9 + b 3 = b T = 3n + 3 n 3 5 7 1st Difference 2 2 2nd Difference not constant constant Solve for
nth term : nth term = a + (n – 1)d1 + ½(n – 1)(n – 2)d2 2 ; 5 ; 10 ; 17 a = 1st term 2 d1 = 1st difference d2 = 2nd difference a d1 3 d2 2 Tn = 2 + (n - 1)3 + ½(n – 1)(n – 2)2 Tn = 2 + 3n - 3 + n - 3n + 2 2 Tn = n + 0n + 1 2 Tn = n + 1 2 T = a.b n n 3 ; 6 ; 12 ; 24 x2 x2 x2 constant ratio use n = 1 : T = a. (2) 1 1 2 2 2 2 b n T 1 3 3 = a.2 3 = a 2 T = 3. (2) n n 2 a b c T = a.b n n constant ratio Sequences & Series Learning Outcomes Solve Arithmetic Sequences Arithmetic Sequence General Term T = a + (n - 1)d n a > first term a d > is the constant difference d n > number of the term(s) n Example: T = a + (n - 1)d n 4 ; 10 ; 16 ; 22 Find the 15th Term simplifly 6 6 6 d constant difference a 4 T = 4 + (n-1)6 n T = 4 + 6n - 6 n T = 6n - 2 n T = 6n - 2 n T = 6n - 2 n n = 15 looking for the 15th term T = 6(15) - 2 15 T = 88 15 WHAT IF THEY ASK IN A DIFFERENT WAY ? Determine the value of n Given Which Term is equal to - 11 19 ; 14 ; 9 ; ..... -5 -5 we know Tn = -11 T = a + (n - 1)d n -11 = a 19 19 + (n-1) d -5 -5 -5 -11 = 19 - 5n + 5 -35 = -5n n = 7 T = -11 7 What if they give
you 2 terms ? Determine the first 3 terms
of an arithmetic sequence
if the 5th term of the sequence
is 12 and the 14th is -33 Write down what you have: T = 12 T = -33 Tn = a + (n -1)d n n have 2 equations 12 = a + (5 - 1)d Tn Tn T = 12 n 12 = a + 4d -33 = a + (14 - 1)d -33 = a + 13d T = -33 n simultaneous
equation subtract 45 = - 9d 0 -5 = d (A) Subsitute d into (A) 12 = a + 4(-5) 32 = a first term 32 + (-5) = 2nd term 27 = T 2 =
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