====== Question 8 and 9, Exercise 4.4 ======
Solutions of Question 8 and 9 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 8=====
Find the next two terms of the geometric sequence:
$$90,30,10 \ldots$$

** Solution. **

Given sequence is geometric with $a_1=90$ and $r=\dfrac{30}{90}=\dfrac{1}{3}$.\\
General term of the geometric series is given as 
$$a_{n}=a_{1} r^{n-1}.$$
Thus
\begin{align*}
& a_{4}=a_{1} r^3=(90)\left(\dfrac{1}{3} \right)^3=90 \times\dfrac{1}{27}=\dfrac{10}{3}\\
& a_{5}=a_{1} r^3=(90)\left(\dfrac{1}{4} \right)^4=90 \times\dfrac{1}{81}=\dfrac{10}{9}\\
\end{align*}
Hence $a_4=\dfrac{10}{3}$, $a_5=\dfrac{10}{9}$. GOOD






=====Question 9=====
Find the next two terms of the geometric sequence: 
$$2,6,18 \ldots$$

** Solution. **

Given sequence is geometric with \(a_1=2\) and \(r=\frac{6}{2}=\frac{3}{1}\).\\
General term of the geometric series is given as 
$a_{n}=a_{1} r^{n-1}.$
Thus
\begin{align*}
& a_{4}=a_{1} r^3=(2)\left(\frac{3}{1}\right)^3=2 \times 27 = 54 \\ 
& a_{5}=a_{1} r^4=(2)\left(\frac{3}{1}\right)^4=2 \times 81 = 162 
\end{align*}
Hence $a_4=54$, $a_5=162$.







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