# 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.

## Question 3

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 \(r = \frac{3}{2}\).

## Question 4

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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