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Question
Which term of the following sequence:
`1/3, 1/9, 1/27`, ...., is `1/19683`?
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Solution
First term of the geometric series a = `1/3`
Second term = `1/9`
∴ Common ratio = `1/9 ÷ 1/3`
= `1/9 xx 3`
= `1/3`
nth term = arn-1
= `1/3 (1/3)^("n" -1)`
= `1/3^"n"`
Given: `1/3^"n"`
= `1/19683`
= `1/3^9`
Hence n = 9
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