====== Question 12 and 13, Exercise 4.4 ======
Solutions of Question 12 and 13 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 12=====
Find the next two terms of the geometric sequence: 
$$\frac{1}{27}, \frac{1}{9}, \frac{1}{3}, \ldots$$

** Solution. **

Given sequence is geometric with \(a_1=\frac{1}{27}\) and \(r=\frac{\frac{1}{9}}{\frac{1}{27}}=3\).\\
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=\left(\frac{1}{27}\right)(3)^3=\frac{1}{27} \times 27 = 1 \\ 
& a_{5}=a_{1} r^4=\left(\frac{1}{27}\right)(3)^4=\frac{1}{27} \times 81 = 3 
\end{align*}
Hence $a_4=1$, $a_5=3$.


=====Question 13=====
Find the next two terms of the geometric sequence: 
$$\frac{1}{4}, \frac{1}{2},-1, \ldots$$

** Solution. **

Given sequence is geometric with \(a_1=\frac{1}{4}\) and \(r=\frac{\frac{1}{2}}{\frac{1}{4}}=2\).\\
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=\left(\frac{1}{4}\right)(2)^3=\frac{1}{4} \times 8 = 2 \\ 
& a_{5}=a_{1} r^4=\left(\frac{1}{4}\right)(2)^4=\frac{1}{4} \times 16 = 4 
\end{align*}
Hence $a_4=2$, $a_5=4$.



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