====== 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.

=====Question 13=====
If $y=\dfrac{x}{3}+\dfrac{x^2}{3^2}+\dfrac{x^3}{3^3}+\ldots$ where $0<x<3$, then show that $x=\dfrac{3 y}{1+y}$.
=====Solution=====
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.

=====Question 14=====
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?
====Solution====
\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}

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