Arithmetic sequences and seriesEdexcel A-Level Maths: Revision notes

An arithmetic sequence has a constant common difference $d$ between consecutive terms: $u_{n+1}=u_n+d$. With first term $a$,

$u_n=a+(n-1)d.$

For $a=7$ and $d=4$: $u_{30}=7+29\times4=123$. The sequence is increasing if $d>0$, decreasing if $d<0$. To find the first term above a target, solve the inequality, e.g. $7+4(n-1)>200$ gives $n>49.25$, so $n=50$.

Information about terms or sums gives equations in $a$ and $d$, which you solve simultaneously. If $u_4=17$ then $a+3d=17$. If $S_{12}=324$ then $6(2a+11d)=324$, so $2a+11d=54$. Doubling the first equation gives $2a+6d=34$; subtracting gives $5d=20$, so $d=4$ and $a=5$.

Always state each equation clearly before solving and check the answer in the original information.

The sum of the first $n$ terms is

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

where $l=a+(n-1)d$ is the last term. Use the second form when you know the first and last terms. For $a=7$, $d=4$, $n=30$: $S_{30}=15(14+116)=1950$. The sum of terms $m$ to $n$ is $S_n-S_{m-1}$, e.g. terms $11$ to $20$ is $S_{20}-S_{10}$.

You must be able to prove $S_n$. Write the sum twice, once reversed:

$S_n=a+(a+d)+\cdots+[a+(n-1)d]$

$S_n=[a+(n-1)d]+[a+(n-2)d]+\cdots+a$

Adding, each of the $n$ pairs equals $2a+(n-1)d$, so $2S_n=n[2a+(n-1)d]$ and $S_n=\frac{n}{2}[2a+(n-1)d]$.

The natural numbers $1,2,3,\ldots,n$ form an arithmetic series with $a=1$, $d=1$ and $l=n$:

$\sum_{r=1}^{n}r=\frac{n(n+1)}{2}.$

For example $1+2+\cdots+40=\frac{40\times41}{2}=820$. In sigma notation, $\sum_{r=1}^{20}(3r+2)=3\times210+2\times20=670$, which is the arithmetic series with $a=5$, $d=3$.

Many questions ask for the number of terms needed, or when a sum changes sign. Form $S_n$ as an expression in $n$, set up an inequality and solve. For $u_n=80-3n$: $a=77$, $d=-3$, $S_n=\frac{n}{2}(157-3n)$. This is negative when $n>52.33$, so $n=53$ is the least value. If $d<0$, the sum is greatest when only positive terms are included: $n=26$, $S_{26}=1027$.