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$\frac{a+2d}{a+d}\le \frac{a+d}{a} \Leftrightarrow \frac{a_{2n+1}}{a_{2n}} \le \frac{a_{2n}}{a_{2n-1}}$
$\frac{ar^2}{ar}\le \frac{ar}{a} \Leftrightarrow \frac{a_{2n+2}}{a_{2n+1}}=\frac{a_{2n+1}}{a_{2n}} \le \frac{a_{2n}}{a_{2n-1}}$
$a_1,a_2,a_3$等差，d>0，$\frac{a_3}{a_2}<\frac{a_2}{a_1}$（不取等） |
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tommywong
发表于 2014-3-26 22:50
