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Question
Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is `1/r^n`.
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Solution
Let the first term of the geometric progression be a and common ratio = `1/"r"^"n"`, then
Sum of n terms = `("a"(1 - "r"^"n"))/(1 - "r")` .....(i)
(n + 1)th term = `"ar"^("n"+ 1 - 1)` = arn
∴ arn + arn + 1 + arn + 2 + ....... up to n terms
= `("ar"^"n"(1 - "r"^"n"))/(1 - "r")` .....(ii)
Dividing equation (i) by (ii), we get
`("Sum of n terms")/("Sum of next n terms") = ("a"(1 - "r"^"n"))/(1 - "r") ÷ ("ar"^ "n"(1 - "r"^"n"))/(1 - "r")`
= `("a"(1 - "r"^"n"))/(1 - "r") xx (1 - "r")/("ar"^"n" (1 - "r"^ "n"))`
= `1/"r"^"n"`
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