geometric progression
回答 (2)
2007-11-24 7:51 am
✔ 最佳答案
Let F be the first term.mth term: a = Fr^(m-1).
nth term: b = Fr^(n-1).
(m+n)th term
= Fr^(m+n-1)
= F[(r^m-1)(r^n-1)(r^1)]
= F[(a/F)(b/F)( r )]
= abr/F.
2007-11-24 7:54 am
A = first term
R = common ratio
T(m) = a = AR^(m-1)
T(n) = b = AR^(n-1)
=> a/b = R^(m-n)
R = (a/b)^[1/(m-n)]
T(m+n) = AR^(m+n-1)
= [A^2 * R^(2m + 2n - 2)]^1/2
= [AR^(m-1) * AR(n-1) * R(m+n)]^1/2
= [a*b* (a/b)^[(m+n)/(m-n)]^1/2
(simplify) = a^(m/(m-n)) * b^(-n/(m-n))
R = common ratio
T(m) = a = AR^(m-1)
T(n) = b = AR^(n-1)
=> a/b = R^(m-n)
R = (a/b)^[1/(m-n)]
T(m+n) = AR^(m+n-1)
= [A^2 * R^(2m + 2n - 2)]^1/2
= [AR^(m-1) * AR(n-1) * R(m+n)]^1/2
= [a*b* (a/b)^[(m+n)/(m-n)]^1/2
(simplify) = a^(m/(m-n)) * b^(-n/(m-n))
收錄日期: 2021-04-13 14:33:43
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