Resources for Arithmetic Sequences
-
Questions
13
With Worked SolutionClick Here -
Video Tutorials
4
Click Here -
HSC Questions
3
With Worked SolutionClick Here
Arithmetic Sequences Theory
![The general term of an arithmetic sequence is \(T_n=a+(n-1) d\) where \(a\) is the first term, \(n\) is the number of terms and \(d\) is the common difference\\ The sum to \(n\) terms of an arithmetic series is of two forms:\\ $\begin{aligned} & S_n=\frac{n}{2}(2 a+(n-1) d) \text { and } \\ & S_n=\frac{n}{2}(a+l) \text { where } l \text { is the last term. } \end{aligned}$\\ \begin{multicols}{2} \textbf{Example}\\ %24641 The fifth term of an arithmetic sequence is 44, and the tenth term is 64 \begin{itemize}[leftmargin=1.7em,nosep] \item[\bf{i)}]Find the value of the first term and the common difference. \item[\bf{ii)}]Find the sum of the first 100 terms. \end{itemize} \hfill \linebreak \textbf{Solution} \begin{itemize}[leftmargin=1.5em,nosep] \item[\bf{i)}] \(\begin{aligned}[t] T_{n} &=a+(n-1) d \\ T_{5} &=a+4 d=44 \\ T_{10} &=a+9 d=64 \\ 5 d &=20 \\ \therefore d &=4 \end{aligned}\)\\ \(\begin{aligned} \text {In (1) } a+4 \times 4 &=44 \\ \therefore a &=28 \end{aligned}\)\\ \(\begin{aligned} &\text{Therefore the first term is 28 and the common} \\&\text{ difference is 4.} \end{aligned}\)\\ \item[\textbf{ii)}] \(\begin{aligned}[t] S_{n} &=\frac{n}{2}(2 a+(n-1) d) \\ S_{100} &=\frac{100}{2}(2 \times 28+99 \times 4) \\ &=22,600 \end{aligned}\) \end{itemize} \end{multicols}](/media/kkhlft2o/4762.png)