====== Question 5, 6 and 7, Exercise 4.4 ======
Solutions of Question 5, 6 and 7 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 5=====
Find the first four terms of the geometric sequence.$a_{1}=3, r=-2$

** Solution. **

Given $a_{1}=3$ and $r=-2$. Use the formula 
$$a_{n}=a_{1} r^{n-1}.$$
Then
\begin{align*}
& a_{2}=a_{1} r=(3)(-2)= -6 \\
& a_{3}=a_{1} r^{2}=(3)(-2)^{2}=3 (4)= 12 \\
& a_{4}=a_{1} r^{3}=(3)(-2)^{3}=3  (-8) = -24
\end{align*}
Hence $a_1=3$, $a_2=-6$, $a_3=12$, $a_4=-24$. GOOD

=====Question 6=====
Find the first four terms of the geometric sequence. $a_{1}=27, r=-\frac{1}{3}$

** Solution. **

Given $a_{1}=27$ and $r=-\frac{1}{3}$. Use the formula 
$$a_{n}=a_{1} r^{n-1}.$$
Then
\begin{align*}
& a_{2}=a_{1} r=(27)\left(-\frac{1}{3}\right) = -9 \\ 
& a_{3}=a_{1} r^{2}=(27)\left(-\frac{1}{3}\right)^{2} = 27 \cdot \frac{1}{9} = 3 \\ 
& a_{4}=a_{1} r^{3}=(27)\left(-\frac{1}{3}\right)^{3} = 27 \cdot \left(-\frac{1}{27}\right) = -1 
\end{align*}
Hence $a_1=27$, $a_2=-9$, $a_3=3$, $a_4=-1$. GOOD

=====Question 7=====
Find the first four terms of the geometric sequence. $\quad a_{1}=12, r=\frac{1}{2}$

** Solution. **

Given $a_{1}=12$ and $r=\frac{1}{2}$. Use the formula 
$$a_{n}=a_{1} r^{n-1}.$$
Then
\begin{align*}
& a_{2}=a_{1} r=(12)\left(\frac{1}{2}\right) = 6 \\ 
& a_{3}=a_{1} r^{2}=(12)\left(\frac{1}{2}\right)^{2} = 12 \cdot \frac{1}{4} = 3 \\ 
& a_{4}=a_{1} r^{3}=(12)\left(\frac{1}{2}\right)^{3} = 12 \cdot \frac{1}{8} = 1.5 
\end{align*}
Hence $a_1=12$, $a_2=6$, $a_3=3$, $a_4=1.5$. GOOD


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