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.

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 $a_4=1$, $a_5=3$.

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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