Show that each one of the following progressions is a G.P. Also,

(i) We have,

$a_{1}=4, a_{2}=-2, a_{3}=1, a_{4}=-\frac{1}{2}$

Now, $\frac{a_{2}}{a_{1}}=\frac{-2}{4}=\frac{-1}{2}, \frac{a_{3}}{a_{2}}=\frac{1}{-2}, \frac{a_{4}}{a_{3}}=\frac{-\frac{1}{2}}{1}=\frac{-1}{2}$

$\therefore \quad \frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\frac{a_{4}}{a_{3}}=\frac{-1}{2}$

Thus, $a_{1}, a_{2}, a_{3}$ and $a_{4}$ are in G.P., where $a=4$ and $r=\frac{-1}{2}$.

(ii) We have,

$a_{1}=\frac{-2}{3}, a_{2}=-6, a_{3}=-54$

Now, $\frac{a_{2}}{a_{1}}=\frac{-6}{\frac{-2}{3}}=9, \frac{a_{3}}{a_{2}}=\frac{-54}{-6}=9$

$\therefore \quad \frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=9$

Thus, $a_{1}, a_{2}$ and $a_{3}$ are in G.P., where $a=\frac{-2}{3}$ and $r=9$.

(iii) We have,

$a_{1}=a, a_{2}=\frac{3 a^{2}}{4}, a_{3}=\frac{9 a^{3}}{16}$

Now, $\frac{a_{2}}{a_{1}}=\frac{\frac{3 a^{2}}{4}}{a}=\frac{3 a}{4}, \frac{a_{3}}{a_{2}}=\frac{\frac{9 a^{3}}{16}}{\frac{3 a^{2}}{4}}=\frac{3 a}{4}$

$\therefore \quad \frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\frac{3 a}{4}$

Thus, $a_{1}, a_{2}$ and $a_{3}$ are in G.P., where the first term is $a$ and the common ratio is $\frac{3 a}{4}$.

(iv) We have,

$a_{1}=\frac{1}{2}, a_{2}=\frac{1}{3}, a_{3}=\frac{2}{9}, a_{4}=\frac{4}{27}$

Now, $\frac{a_{2}}{a_{1}}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}, \frac{a_{3}}{a_{2}}=\frac{\frac{2}{9}}{\frac{1}{3}}=\frac{2}{3}, \frac{a_{4}}{a_{3}}=\frac{\frac{4}{27}}{\frac{2}{9}}=\frac{2}{3}$

$\therefore \frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\frac{a_{4}}{a_{3}}=\frac{2}{3}$

Thus, $a_{1}, a_{2}, a_{3}$ and $a_{4}$ are in G.P., where the first term is $\frac{1}{2}$ and the common ratio is $\frac{2}{3}$.