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Question
Show that one of the following progression is a G.P. Also, find the common ratio in case:
\[a, \frac{3 a^2}{4}, \frac{9 a^3}{16}, . . .\]
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Solution
We have,
\[ a_1 = a , a_2 = \frac{3 a^2}{4}, a_3 = \frac{9 a^3}{16}\]
\[\text { Now, } \frac{a_2}{a_1} = \frac{\frac{3 a^2}{4}}{a} = \frac{3a}{4}, \frac{a_3}{a_2} = \frac{\frac{9 a^3}{16}}{\frac{3 a^2}{4}} = \frac{3a}{4} \]
\[ \therefore \frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{3a}{4}\]
\[\text { Thus, } a_1 , a_2 \text { and } a_3 \text { are in G . P . , where the first term is a and the common ratio is } \frac{3a}{4} .\]
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