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Show that the sequence defined by an = 2/3n, n ∈ N is a G.P. |
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Answer» Given: an = 2/3n Let us consider n = 1, 2, 3, 4, … since n is a natural number. So, a1 = 2/3 a2 = 2/32 = 2/9 a3 = 2/33 = 2/27 a4 = 2/34 = 2/81 In GP, a3/a2 = (2/27) / (2/9) = 2/27 × 9/2 = 1/3 a2/a1 = (2/9) / (2/3) = 2/9 × 3/2 = 1/3 ∴ Common ratio of consecutive term is 1/3. Hence n ∈ N is a G.P. |
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