# Arithmetic Series

A LevelAQAEdexcelOCREdexcel 2022OCR 2022

## Arithmetic Series

A series is a sequence where the goal is to add all the terms together. We will study arithmetic series and geometric series.

Recall: Notation from Sequences:

$a$ is first term

$d$ is difference, the amount we add each time

$n$ is the number of terms in the series

We will also introduce $l$, which is the last term of the series. Since there are $n$ terms in the series and we have an $n$th term equation for arithmetic sequences, we have a formula for $l$:

$l=a+(n-1)d$

Make sure you are happy with the following topics before continuing.

A Level

The sum of an arithmetic series with $n$ terms is:

$S_{n}=\dfrac{n(a+l)}{2}$

We can prove this result:

\begin{aligned}S_{n}&=a+(a+d)+(a+2d)+...+(l-d)+l\\[1.2em]&=l+(l-d)+(l-2d)+...+(a+d)+a\\[1.2em]2S_{n}&=(a+l)+(a+l)+(a+l)+...+(a+l)+(a+l)\\[1.2em]S_{n}&=\dfrac{n(a+l)}{2}\end{aligned}

If we substitute in our formula for $l$, we get:

$S_{n}=\dfrac{n}{2}(2a+(n-1)d)$

A Level

## Sum Notation

$\sum$ means sum, and we can use it instead of writing $S_{n}$ to represent arithmetic series.

Example: $\sum_{n=1}^{20}(3n+4)$ means the sum up to the $20$th term of the arithmetic progression defined by $3n+4$.

A Level

## Natural Number Arithmetic Progressions

The sum of the first $n$ natural numbers (positive whole numbers) is:

$S_n = 1 + 2 + 3 + ... + (n-1) + n$

So $a=1$, $l = n$ and $n = n$.

If we put these values into the formula we have already seen, we would get:

$S_n = \dfrac{1}{2} n (n+1)$

A Level
A Level

## Example 1: Arithmetic Series in Practice

Jon is training for a marathon. Last week his furthest run was $4$ miles. He plans to increase the length he runs by $1.1$ miles per day. What is the total distance he has run by the time he reaches his goal of $26$ miles?

[2 marks]

$a=4$

$d=1.1$

$l=26$

Find number of terms between $a$ and $l$

$n=\dfrac{26-4}{1.1}=\dfrac{22}{1.1}=20$

Need to add $1$ because both the first and last terms are included.

$n=21$

Substitute into formula:

\begin{aligned}S_{n}&=\dfrac{n(a+l)}{2} \\[1.2em] S_{21}&=\dfrac{21(4+26)}{2}\\[1.2em]&= \dfrac{21\times 30}{2}\\[1.2em]&=\dfrac{630}{2}\\[1.2em]&=315 \text{ miles} \end{aligned}

A Level

## Example 2: Sum Notation

Find $\sum_{n=1}^{50}(3n+4)$

[2 marks]

This means the sum of the first $50$ terms of the sequence defined by $3n+4$.

$a=3\times 1+4=3+4=7$

$d=3$

$n=50$

Substitute into formula:

\begin{aligned}\sum_{n=1}^{50}(3n+4)&=\dfrac{50}{2}(2\times 7+(50-1)3)\\[1.2em]&=25(14+49\times 3)\\[1.2em]&=25(14+147)\\[1.2em]&=25\times 161\\[1.2em]&=4025\end{aligned}

A Level

## Example Questions

i) $\sum_{n=1}^{k}n$

ii) Arithmetic progression with:

$a=1$

$d=1$

$n=k$

Sub into formula:

\begin{aligned}\sum_{n=1}^{k}n&=\dfrac{k}{2}(2\times 1+(k-1)1)\\[1.2em]&=\dfrac{k}{2}(2+k-1)\\[1.2em]&=\dfrac{k(k+1)}{2}\end{aligned}

iii) $k=100$

\begin{aligned}\sum_{n=1}^{100}n&=\dfrac{100(100+1)}{2}\\[1.2em]&=\dfrac{100\times 101}{2}\\[1.2em]&=\dfrac{10100}{2}\\[1.2em]&=5050\end{aligned}

iv) $\sum_{n=1}^{k}n=55$

$\dfrac{k(k+1)}{2}=55$

$k(k+1)=110$

$k^{2}+k=110$

$k^{2}+k-110=0$

$(k-10)(k+11)=0$

$k=10$ or $k=-11$

$-11$ not feasible

$k=10$

$a=3$

$d=5$

$n=10$

Substitute into formula:

\begin{aligned}S_{n}&=\dfrac{10}{2}(2\times 3+(10-1)5)\\[1.2em]&=5(6+9\times 5)\\[1.2em]&=5(6+45)\\[1.2em]&=5\times 51\\[1.2em]&=255\end{aligned}

$2$ years is $24$ months.

$a=10$

$d=10$

$n=24$

Substitute into formula:

\begin{aligned}S_{n}&=\dfrac{24}{2}(2\times 10+(24-1)10)\\[1.2em]&=12(20+23\times 10)\\[1.2em]&=12(20+230)\\[1.2em]&=12\times 250\\[1.2em]&=3000\end{aligned}

$S_{n}=184$

$a=9$

$d=4$

Substitute into formula:

$\dfrac{n}{2}(2\times 9+(n-1)4)=184$

$n(18+4n-4)=368$

$n(14+4n)=368$

$4n^{2}+14n=368$

$4n^{2}+14n-368=0$

$2n^{2}+7n-184=0$

$(n-8)(2n+23)=0$

$n=8$ or $n=-\dfrac{23}{2}$

$n=-\dfrac{23}{2}$ not feasible

$n=8$

$\dfrac{1}{2}k (k+1) = 325$

$k^2 + k = 650$

$k^2 + k - 650 = 0$

$(k+26)(k-25) = 0$

$k$ cannot be negative so $k = 25$.

A Level

A Level

A Level

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