Question 7 and 8, Exercise 4.5

Solutions of Question 7 and 8 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. $a_{1}=16, r=-\frac{1}{2}, n=10$

Solution.

Given $a_1 = 16$, $r = -\frac{1}{2}$ and $n = 10$. 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_{10} &= \frac{16 \left(1 - \left(-\frac{1}{2}\right)^{10}\right)}{1 - \left(-\frac{1}{2}\right)} \\ &= \frac{16 \left(1 - \frac{1}{1024}\right)}{1 + \frac{1}{2}} \\ &= \frac{16 \cdot \frac{1023}{1024}}{\frac{3}{2}} \\ &= \frac{32736}{3072} \\ &= 10.66667. \end{align*} Hence, the required sum is approximately $10.67$

Find the sum of the geometric series. $a_{1}=243, r=-\frac{2}{3}, n=5$

Solution.

Given $a_1 = 243$, $r = -\frac{2}{3}$, and $n = 5$.

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_{5} &= \frac{243 \left(1 - \left(-\frac{2}{3}\right)^{5}\right)}{1 - \left(-\frac{2}{3}\right)} \\ &= \frac{243 \left(1 +\frac{32}{243}\right)}{1 + \frac{2}{3}} \\ &= \frac{243 \cdot \frac{275}{243}}{\frac{5}{3}} \\ &= 165. \end{align*} Hence, the required sum is $165$.

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