# Question 13 & 14 Exercise 4.5

Solutions of Question 13 & 14 of Exercise 4.5 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

If $y=\dfrac{x}{3}+\dfrac{x^2}{3^2}+\dfrac{x^3}{3^3}+\ldots$ where $0<x<3$, then show that $x=\dfrac{3 y}{1+y}$.

Adding 1 to both sides of the given series, we get $$1+y=1+\dfrac{x}{3}+\dfrac{x^2}{3^2}+\dfrac{x^3}{3^3}$$ Now the series on R.H.S of the above is geometric series with $a_1=1$,

$r=\dfrac{x}{3}$, and $|r|=\dfrac{x}{3}<1$ because $0<x<3$.

Thus infinite sum exists and is given by $S_{\infty}=\dfrac{a_1}{1-r}$, putting $a_1, \quad r$, we get $$S_{\infty}=\dfrac{1}{1-\dfrac{x}{3}}=\dfrac{3}{3-x}$$ putting in (i), we have \begin{align}1+y&=\dfrac{3}{3-x} \\ \Rightarrow \quad 3-x&=\dfrac{3}{1+y}\\ \Rightarrow \quad x&=3-\dfrac{3}{1+y} \\ \Rightarrow \quad x&=\dfrac{3+3 y-3}{1+y} \\ \Rightarrow \quad x&=\dfrac{3 y}{1+y}\end{align} which is required result.

A ball rebound to half the height from which it is dropped. If it is dropped from $10 \mathrm{ft}$, how far does it travel from the moment it dropped until the moment of its eighth bounce?

\begin{align}S&=10+[10(\dfrac{1}{2})+10(\dfrac{1}{2})]+ \\ & {[10(\dfrac{1}{2})^2+10(\dfrac{1}{2})^2]+\ldots} \\ & +[10(\dfrac{1}{2})^7+10(\dfrac{1}{2})^7] \\ S&=10+2[10(\dfrac{1}{2})+10(\dfrac{1}{2})^2+10(\dfrac{1}{2})^7]...(i)\end{align} The sequence is bracket is geometric sequence, with $a_1=10(\dfrac{1}{2}), r=\dfrac{1}{2}, n=7$.
Then $S_7=\dfrac{10(\dfrac{1}{2})[1-(\dfrac{1}{2})^7]}{1-\dfrac{1}{2}}$ \begin{align}\Rightarrow S_7&=10[1-\dfrac{1}{2^7}] \\ \Rightarrow S_7&=10,0.9921875\\ &=9.921875\end{align} Putting (ii) in (i), we get
\begin{align}S&=10+2(9.921875) \\ & =10+19.84375\\ S&=29.84375\quad \text{ feet approximately}\\ S&=29\dfrac{27}{32}\text{ft}\end{align}