====== Question 26 and 27, Exercise 4.4 ======
Solutions of Question 26 and 27 of Exercise 4.4 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 

=====Question 26=====
A Ping-Pong ball is dropped from a height of $16\,\, ft$ and always rebounds one-fourth of the distance fallen. How high does it rebound the $6$th time?

** Solution. **

Ball dropped from height = $16\,\,ft$

Consider $a_1$, $a_2$, $a_3,...$ represents heights of the rebounds of the ball, then

$$a_1 = 16\times \dfrac{1}{4} = 4\,\, ft.$$

Thus the sequence is geometric with $r=\dfrac{1}{4}$.

We have to find $a_6$, where general term of the geometric series is given as 
$$a_{n}=a_{1} r^{n-1}.$$
Thus
\begin{align*}
a_{6}&=a_{1} r^5 \\
&=(4)\left(\dfrac{1}{4} \right)^5 \\
& = \dfrac{1}{256}
\end{align*}
Hence, the ball will rebound $\dfrac{1}{256}\,\, ft$ at $6$th time.

=====Question 27=====
A city has a current population of 100,000 and the population is increasing by $3 \%$ each year. What will the population be in $15^{\text {th }}$ years?

** Solution. **

Here, 
\begin{align*}
a_1 &=  100,000,  \quad n = 16\\
r &= 1+\frac{3}{100}=1+0.03=1.03
\end{align*}
We have
$$a_n=a_1r^{n-1}$$
Thus
\begin{align*}
a_{16}&= a_1r^{16-1}\\
&=100,000(1.03)^{15}\\
&=100,000(155796.74)\\
&=155,797
\end{align*}
Hence, the population after $15$ years will be $155,797$. GOOD





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