A sequence is a list of numbers arranged in a definite order, each number being a term. The first term is written a₁ or simply a, the second a₂, and the general term aₙ, where the small n gives the position. Sequences appear in laboratory work whenever quantities are stepped by a fixed rule: standard solutions prepared at equally spaced concentrations, an oven temperature raised by the same amount every minute, or reagent volumes added in equal increments.
An arithmetic sequence is a sequence in which each term is found from the previous term by adding the same number. That fixed number is the common difference, d. If the first term is a, the sequence is
a, a + d, a + 2d, a + 3d, ...
The common difference is found by subtracting any term from the one after it: d = a₂ − a₁ = a₃ − a₂. If these differences are all equal, the sequence is arithmetic. If d is positive the terms increase, if d is negative they decrease, and if d = 0 every term is the same.
For example, 12, 17, 22, 27 has differences of 5, 5, 5, so d = 5. The sequence 100, 90, 80, 70 has d = −10. The sequence 2, 4, 8, 16 has differences 2, 4, 8, so it is not arithmetic.
When three numbers a, b, c are consecutive terms of an arithmetic sequence, the middle term b is the arithmetic mean of the other two. Because b − a = c − b, the middle term is half the sum of its neighbours:
b = (a + c)/2
The arithmetic mean of two numbers is the number exactly halfway between them. It is the same calculation as the mean of two data values, described in Mean, used here to find the middle term of a sequence.
To insert m arithmetic means between two numbers a and b, the sequence a, (m means), b has m + 2 terms and m + 1 equal steps. The common difference is therefore
d = (b − a)/(m + 1)
and the means are a + d, a + 2d, ..., a + md.
Example 1: Two consecutive burette readings are 12.4 mL and 13.6 mL. Find the reading halfway between them.
Arithmetic mean = (12.4 + 13.6)/2 = 26.0/2 = 13.0
Answer: 13.0 mL.
Example 2: A calibration run needs standards at equal steps between 0.5 mg/L and 2.5 mg/L. Insert three arithmetic means.
m = 3, so d = (2.5 − 0.5)/(3 + 1) = 2.0/4 = 0.5
The means are 0.5 + 0.5 = 1.0, then 1.5, then 2.0.
The sequence is 0.5, 1.0, 1.5, 2.0, 2.5.
Check: 2.0 + 0.5 = 2.5, which is the last term.
Example 3: A spectrophotometer is to be set at five equally spaced wavelengths between 400 nm and 700 nm. Insert five arithmetic means.
m = 5, so d = (700 − 400)/(5 + 1) = 300/6 = 50
The wavelengths are 450, 500, 550, 600 and 650 nm.
Check: 650 + 50 = 700.
Example 4: The three numbers x, 2x + 1 and 5x − 4 are consecutive terms of an arithmetic sequence. Find x and the common difference.
The middle term is the mean of the other two: 2x + 1 = (x + 5x − 4)/2
4x + 2 = 6x − 4, so 2x = 6 and x = 3
The terms are 3, 7, 11, so d = 4.
The nth term is the term in position n. Each step adds d, so the term in position n is reached by starting at a and adding d a total of (n − 1) times:
aₙ = a + (n − 1)d
For example, the 5th term is a + 4d, because four steps have been taken from the first term.
The formula links four quantities: a, d, n and aₙ. When any three are known, the fourth can be found. When two terms are known, two equations in a and d can be written and solved together.
Example 1: A water bath is heated so that its temperature at the start of each minute is 40, 55, 70, ... °C. Find the temperature at the start of the 10th minute, and the position of the term 250 °C.
a = 40 and d = 55 − 40 = 15
a₁₀ = 40 + (10 − 1)(15) = 40 + 135 = 175
For 250: 250 = 40 + (n − 1)(15), so (n − 1) × 15 = 210, n − 1 = 14 and n = 15
Answer: 175 °C at the start of the 10th minute, and 250 °C is the 15th term.
Example 2: Reagent is added in portions that increase by the same amount each time. The 4th portion is 23 mL and the 9th is 48 mL. Find the first portion and the common difference.
a + 3d = 23 and a + 8d = 48
Subtract: 5d = 25, so d = 5
Then a + 15 = 23, so a = 8
The portions are 8, 13, 18, 23, 28, ... mL.
Check: a₉ = 8 + 8(5) = 48.
Example 3: A stock of 120 mL of reagent is used at 7.5 mL per analysis. The volume remaining after each analysis forms the sequence 112.5, 105, 97.5, ... How many analyses can be done, and what is the 16th term?
a = 112.5 and d = −7.5
aₙ = 112.5 + (n − 1)(−7.5) = 120 − 7.5n
For the stock to be used up, aₙ = 0: 120 − 7.5n = 0, so n = 16
a₁₆ = 120 − 7.5(16) = 120 − 120 = 0
Answer: 16 full analyses can be done, after which none remains. A 17th term would be −7.5 mL, which is not possible, so the sequence stops at the 16th term.
Create a free account to open more of this chapter.
Free: practical guides, quick cards, workplace scenarios and more.
Create a free account