Question 3 and 4, Exercise 4.4

Solutions of Question 3 and 4 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.

Determine whether the given sequence is geometric. If so, find the common ratio. $\frac{3}{2}, \frac{9}{4}, \frac{27}{8}, \frac{81}{16}, \ldots$

Solution. Given sequence is \(\frac{3}{2}, \frac{9}{4}, \frac{27}{8}, \frac{81}{16}, \ldots\).
Suppose \begin{align*} r_1&=\frac{9/4}{3/2} = \frac{9}{4} \times \frac{2}{3} = \frac{3}{2} \\ r_2&=\frac{27/8}{9/4} = \frac{27}{8} \times \frac{4}{9} = \frac{3}{2} \\ r_3&=\frac{81/16}{27/8} = \frac{81}{16} \times \frac{8}{27} = \frac{3}{2} \end{align*} Since $r_1 = r_2 = r_3$, it means two consecutive terms has same ratio.
Hence given sequence is geometric and common ratio \(r = \frac{3}{2}\).

Determine whether each sequence is geometric. If so, find the common ratio. $7,14,21,28, \ldots$

Solution.

Given sequence is $7, 14, 21, 28, \ldots$
Suppose \begin{align*} r_1 &= \frac{14}{7} = 2, \\ r_2 &= \frac{21}{14} = \frac{3}{2}, \\ r_3 &= \frac{28}{21} = \frac{4}{3}. \end{align*} Since $r_1 \neq r_2 \neq r_3,$ it means ratios of consecutive terms are different.
Thus, the sequence $7, 14, 21, 28, \ldots$ is not geometric.

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