Resources for Summing an AP
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Questions
18
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Video Tutorials
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HSC Questions
3
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Summing an AP Theory
![The sum to \(n\) terms of an amithmetic series is of two forms\\ \(\quad S_n=\dfrac{n}{2}(2 a+(n-1) d)\) \\ where \(a\) is the first term, \(n\) is the number of terms and \(d\) is the common difference\\ \(\quad S_n=\dfrac{n}{2}(a+l)\) where \(l\) is the last term.\\ \begin{multicols}{2} \textbf{Example 1}\\ %31044 The sixth term of an arithmetic progression is 18 and the eleventh term is 24. \begin{itemize}[leftmargin=1.7em,nosep] \item[\bf{i)}]Find the common difference \item[\bf{ii)}]Find the sum of the first 20 terms \end{itemize} \textbf{Example 1 solution}\\ \textbf{i)} \(\begin{aligned}[t] T_n &= a + (n-1)d\\T_6 &= a + 5d = 18 & & (1)\\ T_{11} &= a+ 10d = 24 && (2) \end{aligned}\)\\ \(\begin{aligned} (2) - (1) \qquad 5d &= 6\\d &= 1.2 \end{aligned}\)\\ Therefore the common difference is 1.2. \\ \textbf{ii)} \(\begin{aligned}[t] \text{Using } d &= 1.2 \quad \text{(from i)} \end{aligned}\)\\ \(\begin{aligned} \text{In } (1) \quad a+5\times 1.2 &= 18\\a + 6 &= 18\\\therefore a &= 12 \end{aligned}\)\\ \(\begin{aligned} S_n &= \frac{n}{2}(2a + (n-1)d)\\S_{20} &= \frac{20}{2}(2\times 12 + 19\times 1.2) = 468\end{aligned}\)\\ Therefore the sum of the first 20 terms is 468.\\ \columnbreak \textbf{Example 2}\\ %24669 For the first three terms of the series log 2, log 4, log 8 find \begin{itemize}[leftmargin=1.7em,nosep] \item[\bf{i)}]The 6th term \item[\bf{ii)}]The sum of the first six terms. \\ Give your answer in the form \(k\log 2\) \end{itemize} \textbf{Example 2 solution}\\ \(\begin{aligned} \textbf{i)} \;\; &\log 2, \log 4, \log 8, \ldots\\ &\log 2,2 \log 2,3 \log 2\\ &\therefore A P, a=\log 2, d=\log 2\\ &T_{n}=a+(n-1) d\\ &T_{6}=\log 2+5 \log 2\\ &T_{6}=6 \log 2\\ \textbf{ii)}\;\; & \begin{aligned}[t] S_{n} &=\frac{n}{2}(a+l) \\ S_{6} &=\frac{6}{2}\left(\log 2+6 \log 2\right) \\ &=3 \times 7 \log 2 \\ &=21 \log 2 \end{aligned} \end{aligned}\)\\ \end{multicols}](/media/nvchofpw/4765.png)