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

(i) 4, −2, 1, −1/2, ...

(ii) −2/3, −6, −54, ...

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

(iv) 1/2, 1/3, 2/9, 4/27, ...

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Solution

$(i) We have, a_{1} = 4, a_{2} = - 2, a_{3} = 1, a_{4} = - \frac{1}{2} Now, \frac{a_{2}}{a_{1}} = \frac{- 2}{4} = \frac{- 1}{2}, \frac{a_{3}}{a_{2}} = \frac{1}{- 2}, \frac{a_{4}}{a_{3}} = \frac{- \frac{1}{2}}{1} = \frac{- 1}{2} ∴ \frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}} = \frac{a_{4}}{a_{3}} = \frac{- 1}{2} Thus, a_{1}, a_{2}, a_{3} and a_{4} are in G . P ., where a = 4 and r = \frac{- 1}{2} .$

$(ii) We have, a_{1} = \frac{- 2}{3}, a_{2} = - 6, a_{3} = - 54 Now, \frac{a_{2}}{a_{1}} = \frac{- 6}{\frac{- 2}{3}} = 9, \frac{a_{3}}{a_{2}} = \frac{- 54}{- 6} = 9 ∴ \frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}} = 9 Thus, a_{1}, a_{2} and a_{3} are in G . P ., where a = \frac{- 2}{3} and r = 9 .$

$(iii) We have, a_{1} = a, a_{2} = \frac{3 a^{2}}{4}, a_{3} = \frac{9 a^{3}}{16} Now, \frac{a_{2}}{a_{1}} = \frac{\frac{3 a^{2}}{4}}{a} = \frac{3 a}{4}, \frac{a_{3}}{a_{2}} = \frac{\frac{9 a^{3}}{16}}{\frac{3 a^{2}}{4}} = \frac{3 a}{4} ∴ \frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}} = \frac{3 a}{4} Thus, a_{1}, a_{2} and a_{3} are in G . P ., where the first term is a and the common ratio is \frac{3 a}{4} .$

$(iv) We have, a_{1} = \frac{1}{2}, a_{2} = \frac{1}{3}, a_{3} = \frac{2}{9}, a_{4} = \frac{4}{27} Now, \frac{a_{2}}{a_{1}} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}, \frac{a_{3}}{a_{2}} = \frac{\frac{2}{9}}{\frac{1}{3}} = \frac{2}{3}, \frac{a_{4}}{a_{3}} = \frac{\frac{4}{27}}{\frac{2}{9}} = \frac{2}{3} ∴ \frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}} = \frac{a_{4}}{a_{3}} = \frac{2}{3} Thus, a_{1}, a_{2}, a_{3} and a_{4} are in G . P ., where the first term is \frac{1}{2} and the common ratio is \frac{2}{3} .$

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Similar questions

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

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}, . . . . . .

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

(3, 2)

(- 2, 1)

(- 1, - 3)

(5, - 1)

Find :

(i) the ninth term of the G.P. 1, 4, 16, 64, ....

(ii) the 10th term of the G.P. $- \frac{3}{4}, \frac{1}{2}, - \frac{1}{3}, \frac{2}{9}, . . . . . .$

(iii) the 8th term of the G.P. 0.3, 0.06, 0.012, ....

(iv) the 12th term of the G.P. $\frac{1}{a^{3} x^{3}}, a x, a^{5} x^{5},$ .......

(v) nth term of the G.P. $\sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{3 \sqrt{3}}, . . . . .$

(vi) the 10th term of the G.P. $\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}$ , ......

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