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.

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 $S_4=300$.

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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