Show that each one of the following progressions is a G.P. Also, find the common ratio in each case :

(i) $4, - 2, 1, - \frac{1}{2}, . . . . . . .$

(ii) $\frac{- 2}{3} = - 6 = - 54, . . . .$

(iii) $a, \frac{3 a^{2}}{4}, \frac{9 a^{3}}{16}, . . . . . .$

(iv) $\frac{1}{2}, \frac{1}{3}, \frac{2}{9}, \frac{4}{27}, . . . . . . .$

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Solution

(i) 4, - 2, 1, $- \frac{1}{2}, . . . .$

We have,

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

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

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

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

Thus, $a_{1}, a_{2}, a_{3} a n d a_{4}$ are in G.P., where

$a = 4 a n d r = \frac{- 1}{2}$

(ii) $\frac{- 2}{3} = - 6 = - 54, . . . .$

we have,

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

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

$∴ \frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}} = 9$

Thus, $a_{1}, a_{2} a n d a_{3}$ are in G.P., where a

$= \frac{- 2}{3}$ and r = 9.

(iii) $a, \frac{3 a^{2}}{4}, \frac{9 a^{3}}{16}, . . . . .$

We have,

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

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

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

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

(iv) $\frac{1}{2}, \frac{1}{3}, \frac{2}{9}, \frac{4}{27} . . . . . . . . . .$

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

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

$= \frac{\frac{4}{27}}{\frac{2}{9}} = \frac{2}{3}$

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

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

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Similar questions

a, \frac{3 a^{2}}{4}, \frac{9 a^{3}}{16}, . . .

Which term of the G.P. :

(i) $\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}, \frac{1}{4 \sqrt{2}}, . . . . i s \frac{1}{512 \sqrt{2}}$ ?

(ii) 2, $2 \sqrt{2}, 4, . . . . i s 128$ ?

(iii) $\sqrt{3}, 3, 3 \sqrt{3}, . . . . . i s 729$ ?

(iv) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27} . . . i s \frac{1}{19683}$ ?

0.12, 0.24, 0.48, . . . .

0.004, 0.02, 0.1, . . . .

\frac{1}{2}, \frac{1}{3}, \frac{2}{9}, \frac{4}{27}, . . . . .

12, 1, \frac{1}{12}, . . . . .

\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}, . . . .

4, - 2, - 1, - \frac{1}{2}, . . . . . .

\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, 7 \sqrt{2}, . . .

Show that each of the following sequences is an A.P. Also, find the common difference and write 3 more terms in each case.

(i) $3, - 1, - 5, - 9 \dots \dots$

(ii) $- 1, \frac{1}{4}, \frac{3}{2}, \frac{11}{4}, \dots \dots$

(iii) $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, 7 \sqrt{2}, \dots \dots$

(iv) $9, 7, 5, 3, \dots \dots$

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