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Question
Find : `sum_("n" = 1)^oo 0.4^"n"`
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Solution
`sum_("n" = 1)^oo 0.4^"n"`
= 0.4 + (0.4)2 + (0.4)3 + …
The terms 0.4, (0.4)2, (0.4)3 are in G.P.
∴ a = 0.4, r = 0.4
Since, |r| = |0.4| < 1
∴ sum to infinity exists.
∴ `sum_("n" = 1)^oo 0.4^"n" = 0.4/(1 - 0.4)`
= `0.4/0.6`
= `2/3`
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