Arithmetic sequences and seriesEdexcel International A Level Maths: Revision notes

In an arithmetic sequence each term is found by adding a fixed common difference $d$ to the previous term. With first term $a$: $u_n=a+(n-1)d.$ Example: $a=7$, $d=4$ gives $u_{20}=7+19\times4=83$. To find when a term first exceeds a value, solve an inequality in $n$ and round up to an integer: $7+4(n-1)>200\Rightarrow n>49.25$, so $n=50$. Given two terms, form two equations: $u_3=11$ and $u_8=36$ give $a+2d=11$ and $a+7d=36$, so $d=5$, $a=1$.

The sum of the first $n$ terms is $S_n=\frac n2\left[2a+(n-1)d\right]=\frac n2(a+l),$ where $l$ is the last term. Example: $a=7$, $d=4$, $n=20$: $S_{20}=10\times(14+76)=900$. To find the least $n$ for which $S_n$ exceeds a target, form a quadratic inequality, solve it, and check neighbouring integers. For $a=1$, $d=5$, $S_n=\frac{n(5n-3)}{2}>1000$ gives $n=21$ ($S_{20}=970$, $S_{21}=1071$). A sum between two points is a difference: $\sum_{r=11}^{20}u_r=S_{20}-S_{10}$.

The numbers $1,2,3,\dots,n$ form an arithmetic series with $a=1$, $d=1$ and last term $n$: $\sum_{r=1}^{n}r=\frac{n(n+1)}{2}.$ So $1+2+\dots+100=\frac{100\times101}{2}=5050$. Properties of sigma notation: $\sum(u_r+v_r)=\sum u_r+\sum v_r$ and $\sum ku_r=k\sum u_r$, and $\sum_{r=1}^{n}c=nc$. For $u_r=3r+1$: $\sum_{r=1}^{n}(3r+1)=3\cdot\frac{n(n+1)}{2}+n=\frac{n(3n+5)}{2}$.

$\displaystyle\sum_{r=1}^{n}u_r$ means $u_1+u_2+\dots+u_n$: the lower limit is where $r$ starts, the upper limit is where it stops, and $u_r$ is the rule for each term. The number of terms is upper minus lower plus one: $\sum_{r=11}^{20}$ has $10$ terms. If the terms form an arithmetic sequence, use the sum formula; otherwise split the sum using the properties above.

You must know the proof. Write $S_n$ forwards and backwards: $S_n=a+(a+d)+\dots+\left[a+(n-1)d\right]$ $S_n=\left[a+(n-1)d\right]+\dots+(a+d)+a$ Add: each of the $n$ pairs sums to $2a+(n-1)d$, so $2S_n=n\left[2a+(n-1)d\right]$, giving $S_n=\frac n2\left[2a+(n-1)d\right]$.