Arithmetic Series

In an arithmetic series, a set of equally spaced numbers are being added together. The word ‘series‘ indicates that we should be adding the given numbers together. So, instead of seeing a sequence of numbers separated by a comma, they are separated by a plus sign. The word ‘arithmetic‘ indicates that the numbers in the series are increasing (or decreasing) by the same amount each time. We call this the common difference and is usually denoted $d$ . For example, $5 +8+11+14+17+20+23+26+29$ is an arithmetic series with common difference $d=3$ . We refer to the numbers in the series as ‘terms‘. So, the arithmetic series has 9 terms and the first term is 5. We usually denote the number of terms $n$ and the first term $a$ . It follows that the general arithmetic series is as follows:

$a+(a+d)+(a+2d)+(a+3d)+...+a+(n-1)d$

We leave individual terms inside brackets for clarity. Note that the final or $n$ th term in this series is the first term with the common difference added just $n-1$ times. If there are more terms, we may refer to the $n$ th term in any arithmetic series as $u_n=a+(n-1)d$ . You should memorise this. It is also possible to add all of the terms (or just the first $n$ terms) of any arithmetic series using the formula

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

The ‘S’ stands for ‘sum’. This formula is given in the Edexcel formula booklet. See below for proof of the formula and see Examples 1 to 3 for some arithmetic series and their summations. An alternative expression is given as $S_n=\frac{n}{2}\left[a+l\right]$ if the last time $l$ is known instead of $d$ .

Proof of the summation formula for Arithmetic Series

The proof of the formula is started off by writing out $S_n$ so that the $n$ terms are visible. The … indicates that there are some terms in between that follow the pattern as expected. We then write it out again in reverse order:

$S_n=[a]+(a+d)+(a+2d)+(a+3d)+...+\left\lbrace a+(n-1)d\right\rbrace$

$S_n=[a+(n-1)d]+...+(a+3d)+(a+2d)+(a+d)+\left\lbrace a\right\rbrace$

We now add the two expressions together. Note that we are doing this in pairs, as shown for example, by the square and curly brackets:

$2S_n=[2a+(n-1)d]+2a+(n-1)d+...+\left\lbrace 2a+(n-1)d\right\rbrace=n\left[2a+(n-1)d\right]$

since each of the pairs adds to make $2a+(n-1)d$ and there are $n$ of them. It follows that $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$ by dividing both sides of the last expression by 2.

Sigma Notation

There is a shorthand way of writing out the terms in an arithmetic series – we can use sigma notation. The symbol $\sum$ is the Greek letter capital ‘sigma’ and in mathematics it indicates a summation. There is usually an index=number along the bottom, another number on the top and an expression to the right of the sigma symbol. This tells us to first set the index to the bottom number and insert it into the expression. Do this again for each integer between the top and bottom numbers and add them all together. For example,

$\sum_{r=1}^{7}(2r+5)=7+9+11+13+15+17+19=91$ .

The expression to the right of sigma is $2r+5$ . The index is $r$ and we start by putting $r=1$ into the expression to get 7. We then put $r=2$ to get 9 and so on until the last term where we set $r=7$ to get 19. These terms are all added together to get a result of 91. The $S_n$ formula can be used to evaluate arithmetic series written using sigma notation. See Example 4. Note that the index does not have to start from 1. Also note that the expression may not depend on the index. For example, $\sum_{i=1}^51=1+1+1+1+1=5$ . In general, $\sum_{i=1}^n1=n$ .

We can also use the sigma notation for other summations such as the first 5 square numbers, for instance. We can write this as $\sum_{j=1}^5 j^2=1^2+2^2+3^2+4^2+5^2$ . However, we can’t use the arithmetic series results from above to evaluate this.

Examples

Find the 20th term and the sum of the first 20 terms for the following arithmetic series:

$-6-1+4+9+14+…$

Solution:

The first term is $a=-6$ and the common difference is $d=5$ . Hence, the 20th term is $u_{20}=-6+5(20-1)=89$ and the sum of the first 20 terms is:

$S_{20}=\frac{20}{2}\left(-12+5(20-1)\right)=830$

Consider the arithmetic sequence with $n$ th term $u_n=311-7n$ . Find the 100th term of the sequence. Find the first negative term in the sequence.

Solution:

The 100th term in the sequence is found by inserting $n=100$ :

$u_{100}=311-7\times 100=-389$ .

The first few terms in the sequence are 304, 297, 290, … since the common difference is -7 and so at some point the terms will become negative. By solving $311-7n<0$ , i.e. $n>44.43$ to two decimal places. Hence, the first negative term is the 45th term:

$u_{45}=311-7\times 45=-4$ .

Find the number of terms required for the sum of the series with $n$ th term $u_n=3+4(n-1)$ to exceed 190.

Solution:

We are looking for the smallest $n$ such that $S_n=\frac{n}{2}\left[2a+(n-1)d\right]>190$ where the first term $a=3$ and the common difference $d=4$ :

$S_n=\frac{n}{2}\left[6+4(n-1)\right]>190$

Multiplying out gives $3n +2n(n-1)>190$ or $2n^2+n-190>0$ . Factorising gives $(2n-19)(n+10)>0$ and so $n<-10$ or $n>9.5$ . See more on solving quadratic inequalities. It follows that $n>9.5$ and so 10 terms are required for the summation to exceed 190. (This is easily checked by adding up the first 9 and 10 terms since $n$ is relatively small).

Evaluate $\sum_{k=3}^{42}(6k-5)$ .

Solution:

Notice that the first index number is 3, so we start by inserting $k=3$ into the expression giving 13. Inserting $k=4$ gives 19 and the sequence is linear. Hence,

$\sum_{k=3}^{42}(6k-5)=13+19+25+…+247$

where we obtain the last term by setting $k=42$ . There are 40 terms in this sequence, only $k=1$ and $k=2$ are omitted. Hence, we can use the $S_n$ formula noting that $a=13$ , $d=6$ and $n=40$ :