====== Question 1 Exercise 4.4 ======
Solutions of Question 1 of Exercise 4.4 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

=====Question(i)=====
Write the first five terms of geometric ric sequence given that $a_1=5, \quad r=3$
====Solution====
The gcometric sequence is $a_1, a_1 r, a_1 r^2, a_1 r^3, a_1 r^4, \ldots$, so for $a_1=5 ; r=3$, we have
\begin{align}&5,5.3,5.3^2, 5.3^3, 5.3^4, \ldots\\
\Rightarrow &5,15,45,135,405, \ldots\end{align}

=====Question(ii)=====
Write the first five terms of geometric ric sequence given that $a_1=8, \quad r=-\dfrac{1}{2}$
====Solution====
The geomerric sequence is $a_1, a_1 r, a_1 r^2, a_1 r^3, a_1 r^4, \ldots$,\\
so for $$a_1=8 ; r=-\dfrac{1}{2}$$
we have\\
\begin{align}&8,8(-\dfrac{1}{2}), 8(-\dfrac{1}{2})^2, 8(-\dfrac{1}{2})^3,8(-\dfrac{1}{2})^4, \ldots\\
\Rightarrow &8,-4,2,-1, \dfrac{1}{2}, \ldots\end{align}

=====Question(iii)=====
Write the first five terms of geometric ric sequence given that $a_1=-\dfrac{9}{16}, \quad r=-\dfrac{2}{3}$
====Solution====
The geometric sequence is $a_1, a_1 r, a_1 r^2, a_1 r^3, a_1 r^4, \ldots$\\
so for $$a_1=-\dfrac{9}{16} ; r=-\dfrac{2}{3}$$ 
we have\\
\begin{align}&-\dfrac{9}{16} ,-\dfrac{9}{16}(-\dfrac{2}{3}),-\dfrac{9}{16}(-\dfrac{2}{3})^2,\\
&-\dfrac{9}{16} \cdot(-\dfrac{2}{3})^3,-\dfrac{9}{16}(-\dfrac{2}{3})^4, \ldots\\
\Rightarrow&-\dfrac{9}{16}, \dfrac{3}{8},-\dfrac{1}{4}, \dfrac{1}{6},-\dfrac{1}{9},...\end{align}

=====Question(iv)=====
Write the first five terms of geometric ric sequence given that $a_1=\dfrac{x}{y}, \quad r=-\dfrac{y}{x}$
====Solution====
The geometric sequence is, $a_1, a_1 r, a_1 r^2, a_1 r^3, a_1 r^4, \ldots$,\\
so for $$a_1=\dfrac{x}{y} r=-\dfrac{y}{x}$$
we have,\\
\begin{align}&\dfrac{x}{y}, \dfrac{x}{y} \cdot(-\dfrac{y}{x}), \dfrac{x}{y}(-\dfrac{y}{x})^2, \dfrac{x}{y}(-\dfrac{y}{x})^3,\dfrac{x}{y}(-\dfrac{y}{x})^4, \ldots \\
\Rightarrow &\dfrac{x}{y},-1, \dfrac{y}{x},-\dfrac{y^2}{x^2}, \dfrac{y^3}{x^3}, \ldots\end{align}





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