Arithmetic Series (AQA A-Level Mathematics): Revision Notes

4.3.2 Arithmetic Series

Arithmetic Series

The word 'series' describes the act of adding all of the terms in a sequence together.

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Formulae for Arithmetic Series

$n^{th}$ term:

$$u_1 = a$$ $$u_2 = a + d$$ $$u_3 = a + 2d$$ $$u_n = a + (n-1)d$$

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Summing Series

$S_n$ denotes summing the first $n$ terms of a series.

e.g. for the sequence $1, 7, 13, 19, 25$

$S_3$ is the sum of the first $3$ terms $\Rightarrow S_3 = 1 + 7 + 13 = 21$

$$S_n = \frac{n}{2} [2a + (n-1)d]$$

where $n$ is the position of the last term, $a$ is the first term, and $d$ is the common difference.

Proof:

$$S_n = a + (a + d) + (a + 2d) + (a + 3d) + \ldots + [a + (n-1)d]$$ $$= na + d + 2d + 3d + \ldots + (n-1)d$$

$S_n$ in reverse is $S_n = [(n-1)d] + [a + (n-2)d] + \ldots + 3d + 2d + d + na$

Adding $S_n$ to its reversed version:

$$S_n + S_n = 2S_n = na + d + 2d + 3d + \ldots + (n-1)d$$

• $na + (n-1)d + (n-2)d + \ldots + d$ Grouping Corresponding terms

$$= 2S_n = 2na + nd + nd + nd + \ldots + nd \quad \text{(n-1) of these}$$ $$\Rightarrow 2S_n = 2na + (n-1)nd$$ $$\Rightarrow 2S_n = n[2a + (n-1)d]$$ $$\Rightarrow S_n = \frac{n}{2} [2a + (n-1)d]$$

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The third and eighth terms of an arithmetic series are 72 and 37, respectively.

a. Find the first term and common difference of the series.

b. Find the sum of the first 25 terms of the series.

• From the given information, we can set up the equations:

$$a + 2d = 72 \quad \text{(Equation 1)}\\ a + 7d = 37 \quad \text{(Equation 2)}$$

Extract information from the question and write in terms of $a$ and $d$.

• Solving these equations:

$$\text{From the calculator:} \quad :success[a = 86], \quad :success[d = -7]$$

(Make use of calculators if full working is not required.)

b. To find the sum of the first $25$ terms $S_{25}$:

$$S_{25} = \frac{1}{2} (25) [2(86) + 24(-7)] = :success[50]$$

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Summing series in which the first term is not $n = 1$

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