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Write the G.P. whose 4th term is 54 and 7th term is 1458. |
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Answer» t4 = ar3 = 54 ...(i) t7 = ar6 = 1458 ...(ii) ∴ Dividing (ii) by (i), we get \(\frac{t_7}{t_4}\) = r3 = \(\frac{1458}{54}\) = 27 ⇒ r = 3 ∴ From (i), a.(3)3 = 54 ⇒ 27a = 54 ⇒ a = 2 ∴ The G.P. is 2, 2 x 3, 2 x 32, ..... , i.e. 2, 6, 18, 54, .... . |
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