Arithmetic Progressions If you have the sequence 2, 8, 14, 20, 26, then each term is 6 more than the previous term. This is an example of an arithmetic progression (AP) and the constant value that defines the difference between any two consecutive terms is called the common difference. If an arithmetic difference has a first term a and a common difference of d, then we can write a, (a + d), (a + 2d),... {a + (n-1) d}where the nth term = a + (n−1)dSum of Arithmetic seriesThe sum of an arithmetic series of n terms is found by making n/2 pairs each with the value of the sum of the first and last term. (Try this with the sum of the first 10 integers, by making 5 pairs of 11.)This gives us the formula: where a = first term and l = last term. As the last term is the nth term = a + (n − 1)d we can rewrite this as:(Use the first formula if you know the first and last terms; use the second if you know the first term and the common difference.)Geometric ProgressionsIf you have a sequence such as: 81, 27, 9, 3, 1, 1/3, 1/9,... then each term is one third of the term before. This can be written as 81, 81(1/3), 81(1/3)2, 81(1/3)3, 81(1/3)4,... It is an example of a Geometric Progression (GP) where the each term is a multiple of the previous one. The multiplying factor is called the common ratio. So a GP with a first term a and a common ratio r with n terms, can be stated as a, ar, ar2, ar3, ar4... arn-1 , where the nth term = arn-1 Example: In the sequence, 400, 200, 100, 50,... find the 8th term. a = 400, r = 0.5 and so the 8th term = 400 × 0.57 = 3.125Note: To find which term has a certain value you will need to use logarithms. Example: In the sequence, 2, 6, 18, 54 ... which is the first term to exceed 1,000,000? a = 2, r = 3.2 × 3n-1 > 1,000,000 3n-1 > 500000(n − 1) log 3 > log 500000n > 12.94 Therefore: n = 13 Example: In the earlier sequence, 400, 200, 100, 50 ... which is the first term that is less than 1?400 × 0.5(n-1] < 1 0.5(n-1) < 0.0025(n-1) log 0.5 < log 0.0025Therefore: n > 9, or n = 10Note: The inequality sign changed because we divided by a negative (log 0.5 < 0)Sum of Geometric seriesThe sum of the terms can be written in two ways. where a = first term, r = common ratio and r ≠ 1. (use this formula when r < 1). Example: Evaluate,(Note: there are 9 terms.) The first term is when n = 2(i. e 2.362 = 5.5696)Using the formula for the sum of a geometric progression gives: which is approximately 9300 (to 3 s. f.). ConvergenceThe sum of an infinite series exists if:-1 < r < 1 or | r | < 1This is because each successive term is getting smaller and so the series will tend towards a certain limit. This limit is found using the second of our two formulae: If | r | < 1 then as n → ∞, rn → 0 and so: Example: the series 1/3 + (1/3)2 + (1/3)3 + (1/3)4 + ... converges and its sum is 1 as n approaches ∞. (A sequence such as n3 has the first 6 terms as 1 + 8 + 27 + 64 + 125 + 216. As n approaches infinity, the sum also increases. Therefore, it is not convergent. This series is divergent. Every AP has a sum that approaches infinity as n increases, so every AP is divergent.)ExampleFind 1 - 1/2 + 1/4 - 1/8 + ...1 - 1/2 + 1/4 - 1/8 + ... = 1 + (-1/2) + (-1/2)2 + (-1/2)3 + ... This is a geometric progression where r = -½, so | r | < 1. Therefore this series converges to: Two final pieces of information that may be useful: Arithmetic meanThe arithmetic mean of two numbers m and n is given by: Arithmetic mean = ½(m+n)This is the way of finding a missing term in between two known terms. Example: The 4th term of an AP is 14, the 6th term is 22. The 5th term will be the Arithmetic Mean of these two values. i. e. (14 + 22)/2 = 18 (here d = 4 and a = 2). Geometric meanThe geometric mean of two numbers m and n is given by: Geometric mean = √(mn)This represents the value between two others in a GP. Example: The 7th term of a GP is 6, the 9th is 1.5. The 8th term is: √(6×1.5) = √9 = 3Here r = 0.5 and a = 384.