Question 4 & 5 Exercise 5.2

Solutions of Question 4 & 5 of Exercise 5.2 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Find the sum of the following arithtical geometrical series $5+\dfrac{7}{3}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots$

Let $$\begin{align} & S_{\infty}=5+\dfrac{7}{3}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots.(i) \\ & \dfrac{1}{3} S_{\infty}=\dfrac{5}{3}+\dfrac{7}{3^2}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots.(ii) \end{align}$$ Subtracting the (ii) from (i), we get $$\begin{align} & \dfrac{2}{3} S_{\infty}=5+\dfrac{2}{3}+\dfrac{2}{3^2}+\dfrac{2}{3^3}+\ldots \\ & =5+2(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\ldots) \\ & =5+2 \dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}\end{align}$$ $\because$ the series is geometric $r=\dfrac{1}{3}<1$ $$\begin{align}& =5+2 \cdot \dfrac{1}{3} \cdot \dfrac{3}{2} \\ & =5 \cdot \dfrac{3}{2}+\dfrac{3}{2}=\dfrac{18}{2}=9 \\ & \Rightarrow S_{\infty}=9\end{align}$$

If the sum to infinity of the series $3+5 r+7 r^2+\ldots \infty$ is $\dfrac{44}{9}$, find the value of $r$.

Let $$S_{\infty}=3+5 r+7 r^2+\ldots \infty....(i)$$ $$r S_{\infty}=3 r+5 r^2+7 r^3+\ldots \infty....(ii)$$

Subtracsing the (ii) from (i),

we get $$\begin{align} (1-r) S_{\infty}&=3+2 r+2 r^2+2 r^3+\ldots \infty\\ & =3+2 \cdot[r+r^2+r^3+\ldots \infty] \\ & =3+2 \cdot \dfrac{r}{1-r} \\ \Rightarrow S_{\infty}&=3+\dfrac{2 r}{1-r}\end{align}$$ But we are given $S_{\infty}=\dfrac{44}{9}$ $$\begin{align} & \therefore 3+\dfrac{2 r}{1-r}=\dfrac{44}{9} \\ & \Rightarrow \quad \dfrac{2 r}{1-r}=\dfrac{44}{9}-3=\dfrac{17}{9} \\ & \Rightarrow 18 r=17-17 r \\ & \Rightarrow 18 r+17 r=17 \\ & \Rightarrow 35 r=17 \\ & \Rightarrow r=\dfrac{17}{35} \end{align}$$

< Question 2 & 3