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| AM and GM equality (Posted on 2008-01-23) |
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If Arithmetic Mean of 1st k terms of Arithmetic Progression with 1st term a and common difference d is equal to the Geometric Mean of 1st k terms of Geometric Progression with 1st term a and common ratio r, then find the range of r for which a≈2d/(r-1).
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Submitted by Praneeth
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Rating: 5.0000 (1 votes)
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Solution:
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AM = (1/k)*{(k/2)*(2a+(k-1)d)} = (2a+(k-1)d)/2
GM=(akrk*(k-1)/2)1/k
=ar(k-1)/2
=> 2a = (k-1)*d/(r(k-1)/2-1)
If 2>r>0 => 1 > (r-1) > -1
(1+(r-1))(k-1)/2≈1+(k-1)*(r-1)/2+(k-1)*(k-3)*(r-1)/4
=> 2a≈(k-1)*d*4/(k-1)*(r-1){2+(k-3)*(r-1)}
=> For the given condition to satisfy,
2+(k-3)(r-1)≈1
(r-1)≈1/(3-k)
The conditions are as follows:
3>k>-3 and r ≈ 1/(3-k)
For |k|<3, 1/(3-k)+x > r > 1/(3-k)-x where x is positive real very near to 0. |
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