NSW Y12 Maths - Advanced Sequences and Series Summing a GP

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Summing a GP Theory

The sum to \(n\) terms of a geometric series is of two forms\\  \(\quad S_n=\dfrac{a\left(r^n-1\right)}{r-1}\), for \(r>1\), \\  where \(a\) is the first term, \(n\) is the number of terms and \(r\) is the common ratio.\\  \(\quad S_n=\dfrac{a\left(1-r^n\right)}{1-r} \text { for } r<1\)\\  \begin{multicols}{2}  \textbf{Example}\\%31055 A geometric series has a second term 4.5 and the ratio of the fourth term to the third term is 1.5. \begin{itemize} \item[\bf{i)}]Find the common ratio \(r\). \item[\bf{ii)}]What is the first term \(a\)? \item[\bf{iii)}]Calculate the sum of the first 9 terms. \end{itemize} \vfill  \columnbreak  \textbf{Solution}\\ i) \\ \[r = \frac{T_n}{T_{n-1}} =\frac{T_4}{T_3} = 1.5\] ii) \\ $\begin{aligned} T_n &= ar^{n-1}\\T_2 &= ar = 4.5\\a\times 1.5 &= 4.5\\\therefore \text{ the first term is } 3 \end{aligned}$\\ iii) \\ $\begin{aligned}S_n &= \frac{a(r^n - 1)}{r-1}\\S_9 &= \frac{3\left((1.5)^9 - 1\right)}{1.5-1}\\&= 6\left(\left(\frac{3}{2}\right)^9 - 1 \right)\\&= 6\left(\frac{19683}{512} - 1\right)\\&= 224\frac{169}{256}\end{aligned}$\\  \end{multicols}

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