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Question
Find the sum to n terms of the sequence.
0.2, 0.02, 0.002, ...
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Solution
Here, t1 = 0.2, t2 = 0.02, t3 = 0.002
∴ `"t"_2/"t"_1 = 0.02/0.2 = 0.1` and `"t"_3/"t"_2 = 0.002/0.02 = 0.1`
∴ The given sequence is a G.P.
∴ a = 0.2 and r = 0.1
∴ Sn = `("a"(1 - "r"^"n"))/(1 - "r")` for r < 1
= `(0.2[1 - (0.1)^"n"])/(1 - 0.1)`
`= 0.2/0.9 [1 - (0.1)^"n"]`
= `2/9[1 - (1/10)^"n"]`
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