Question 9 and 10, Exercise 4.5
Solutions of Question 9 and 10 of Exercise 4.5 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 9
Find the sum of the geometric series. $a_{1}=343, a_{4}=-1, r=-\frac{1}{7}$
Solution.
Given $a_{1}=343$, $a_{4}=-1$, $r=-\frac{1}{7}$
Let $S_n$ represents the sum of geometric series. Then
$$ S_n =\frac{a_1-a_n r}{1-r}, \quad r\neq 1.$$
Thus
\begin{align*}
S_4 & =\frac{343-(-1)\left(-\frac{1}{7}\right)}{1+\frac{1}{7}} \\
&=\frac{\frac{2400}{7}}{\frac{8}{7}} \\
&=300.
\end{align*}
Hence $S_4=300$. 
Question 10
Find the sum of the geometric series. $a_{3}=\frac{3}{4}, a_{6}=\frac{3}{32}, n=6$
Solution.
Given $a_3 = \frac{3}{4}$, $a_6 = \frac{3}{32}$ and $n = 6$.
Let $a_1$ be first term and $r$ be common ratio, then general term of geometric series is given as
$$a_n=a_1 r^{n-1}$$.
Take
\begin{align*}
&\frac{a_6}{a_3}=\frac{3/32}{3/4} \\
\implies & \frac{a_1 r^5}{a_1 r^2}=\frac{4}{32}\\
\implies & r^3=\frac{1}{8} \\
\implies & r^3=\left(\frac{1}{2} \right)^3\\
\implies & r=\frac{1}{2}.
\end{align*}
Now
\begin{align*}
& a_3 = a_1 r^{2}\\
\implies &\frac{3}{4} = a_1 \left(\frac{1}{2}\right)^{2} \\
\implies &\frac{3}{4} = a_1 \frac{1}{4}\\
\implies a_1 = 3.
\end{align*}
The formula to find the sum of $n$ terms of a geometric series is
$$S_n = \frac{a_1 \left(1 - r^n\right)}{1 - r}, \quad r \neq 1.$$
Thus,
\begin{align*}
S_6 &= \frac{3 \left(1 - \left(\frac{1}{2}\right)^{6}\right)}{1 - \frac{1}{2}} \\
&= \frac{3 \left(1 - \frac{1}{64}\right)}{\frac{1}{2}} \\
&= \frac{3 \cdot \frac{63}{64}}{\frac{1}{2}} \\
&= \frac{189}{32}
\end{align*}
Hence, the required sum is $\dfrac{189}{32}$. 
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