====== Question 18 and 19, Exercise 4.4 ======
Solutions of Question 18 and 19 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 18=====
Find the nth term of the geometric sequence if $a_{1}=32, n=6, r=-\frac{1}{2}$

** Solution. **

Given $a_{1}=32$, $n=6$, $r=-\frac{1}{2}$.\\
General term of the geometric series is given by
$a_{n}=a_{1} r^{n-1}.$
Thus
\begin{align*}
a_6 &= 32 \times \left(-\frac{1}{2}\right)^{6-1} \\ 
&= 32 \times \left(-\frac{1}{2}\right)^{5} \\ 
&= 32 \times \left(-\frac{1}{32}\right) \\ 
&= -1.
\end{align*}
Hence $a_6=-1$.

=====Question 19=====
Find the nth term of the geometric sequence if $a_{1}=16, n=8, r=\frac{1}{2}$

** Solution. **

Given $a_{1}=16$, $n=8$, $r=\frac{1}{2}$.\\
General term of the geometric series is given by
$a_{n}=a_{1} r^{n-1}.$
Thus
\begin{align*}
a_8 &= 16 \times \left(\frac{1}{2}\right)^{8-1} \\ 
&= 16 \times \left(\frac{1}{2}\right)^{7} \\ 
&= 16 \times \frac{1}{128} \\ 
&= \frac{16}{128} = \frac{1}{8}.
\end{align*}
Hence $a_8=\frac{1}{8}$.

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