InterviewSolution
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InterviewSolution
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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, -54iii. a, \(\frac{3a^2}{4}, \frac{9a^3}{16}\), .....iv. \(\frac{1}{2}, \frac{1}{3}, \frac{2}{9}, \frac{4}{27}\), ...... |
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Answer» (i) Let a = 4, b = -2, c = 1. In GP, b2= ac ⇒ (-2)2 = 4.1 ⇒ 4 = 4 Common ratio = r = \(\frac{-2}{4}\) = \(\frac{-1}{2}\) (ii) Let a = \(\frac{-2}{3}\) , b = -6, c = -54. In GP, b2= ac ⇒ (-6)2 = \(\frac{-2}{3}\) x -54 ⇒ 36 = 36 Common ratio = r = \(\cfrac{-6}{\frac{-2}{3}}\) = 9 (iii) Let a = a, b = \(\frac{3a^2}{4}, \) C = \(\frac{9a^2}{16}\) In GP, b2 = ac \(\Rightarrow\) \((\frac{3a^2}{4})^2\) = \(\frac{9a^2}{16} \times a\) \(\Rightarrow\) \(\frac{9a^4}{4} = \frac{9a^4}{16}\) Common ratio = r = \(\frac{\frac{3a^2}{4}}{a}\) = \(\frac{3a}{4}\) (iv) Let a = \(\frac{1}{2}\), b = \(\frac{1}{3}\), c = \(\frac{2}{9}\) In GP, b2 = ac \(\Rightarrow\) \((\frac{1}{3})^2\) = \(\frac{1}{2} \times \frac{2}{9}\) \(\Rightarrow\) \(\frac{1}{9} = \frac{1}{9}\) Common ratio = r = \(\cfrac{\frac{1}{3}}{\frac{1}{2}}\) = \(\frac{2}{3}\) |
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