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9.4A. Infinite Sequence An infinite sequence is a function whose domain is the set of positive integers. Ex. f(n) = 5n + 1 nf(n) 16 211 316 421 526.

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Presentation on theme: "9.4A. Infinite Sequence An infinite sequence is a function whose domain is the set of positive integers. Ex. f(n) = 5n + 1 nf(n) 16 211 316 421 526."— Presentation transcript:

1 9.4A

2 Infinite Sequence An infinite sequence is a function whose domain is the set of positive integers. Ex. f(n) = 5n + 1 nf(n) 16 211 316 421 526

3 Terms of a Sequence The letter ‘a’ is used to represent sequential functions. The functional value of ‘a’ at ‘n’ is written a n (read as “a sub n”). The sequence is expressed as a 1, a 2, a 3, a 4, etc. Ex. Find the first five terms of the sequence a n = 2n – 3 a 1 = 2(1) – 3 = -1 a 2 = 2(2) – 3 = 1 a 3 = 2(3) – 3 = 3 a 4 = 2(4) – 3 = 5 a 5 = 2(5) – 3 = 7

4 Arithmetic Sequence An arithmetic sequence is a sequence that has a common difference between successive terms. Ex. 1, 8, 15, 22, 29 common difference = 7 Ex. 4, 7, 10, 13, 16 common difference = 3 a 1, a 2, a 3, a 4 … is an arithmetic sequence if and only if there is a real number ‘d’ such that a k+1 – a k = d for every positive integer k, where d is the common difference. First term: a 1 Second term: a 1 + d Third term: a 1 + 2d Fourth term: a 1 + 3d nth term: a 1 + (n – 1)d

5 The General Term of an Arithmetic Sequence The General Term of an Arithmetic Sequence is given by a n = a 1 + (n – 1)d Ex. 6, 2, -2, -6 The common difference is -4, so d = -4 The First Term is 6, so a 1 = 6 a n = a 1 + (n – 1)d a n = 6 + (n – 1)(-4) a n = 6 – 4n + 4 a n = -4n + 10

6 Find a specific term of a sequence Ex. Find the 40 th term of the arithmetic sequence 1, 5, 9, 13… Common difference = 4 First term = 1 a n = a 1 + (n – 1)d a n = 1 + (n – 1)(4) a n = 1 + 4n – 4 a n = 4n – 3 a 40 = 4(40) – 3 = 160 – 3 = 157

7 Find the first term of a sequence Find the first term of an arithmetic sequence where the fourth term is 26 and the ninth term is 61. 26 = a 1 + (4 – 1)d = a 1 + 3d 61 = a 1 + (9 – 1)d = a 1 + 8d a 1 + 3d = 26a 1 + 3(7) = 26 a 1 + 8d = 61a 1 + 21 = 26 -5d = -35a 1 = 5 d = 7


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