浙江 2020 · 数学 q07

题面

7．已知等差数列$\left\{a_{n} \right\}$的前n项和S_n，公差d≠0，$\frac{a_{1}}{d}{\leqslant} 1$．记b_{1} = S_{2}$$b_{n + 1} = S_{n + 2} - S_{2 n}，n∈N*，下列等式不可能成立的是（ ） A$2 a_{4}{=} a_{2}{+} a_{6}$ B．$2 b_{4} = b_{2} + b_{6}$ C$a_{4}{}^{2} = a_{2} a_{8}$ D$b_{4}{}^{2} = b_{2} b_{8}$

解析

分析

由已知利用等差数列的通项公式判断 A 与 C；由数列递推式分别求得b_{2} , \ b_{4}$$b_{6} , \ b_{8} ,，分析B，D成立时是否满足公差$d {\neq} 0 , \ {\frac{a_{1}}{d}}{\leqslant} 1$判断B与D

解：在等差数列$\left\{a_{n} \right\}$中，a_{n} = a_{1} + ( n - 1 ) d ,$$\mathrm S_{n + 2} \mathrm{= ( n + 2 ) \ n_{1} + \frac{\ l ( n + 2 ) \ ( \mathrm{n + 1} ) \ l ( \mathrm{n + 2} ) \ l ( \mathrm{n + 1} ) \ l ( \mathrm{2 n - 1} )}{2} d} , \mathrm S_{2 n} \mathrm{= 2 n a_{1} + \frac{\ l^{2 n ( 2 n - 1 )}}{2} d} ,$$b_{1} = S_{2} = 2 a_{1} + d , b_{n + 1} = S_{n + 2} - S_{2 n} = \left( 2 - n \right) {\mathrm{a_{1}}} - \frac{3 {\mathrm{n}}^{2} - 5 {\mathrm{n}} - 2}{2} \mathrm{d} .$$\therefore b_{2} = a_{1} + 2 d , b_{4} = - a_{1} - 5 d , b_{6} = - 3 a_{1} - 2 4 d , b_{8} = - 5 a_{1} - 5 5 d .$$A . 2 a_{4} = 2 ( a_{1} + 3 d ) = 2 a_{1} + 6 d , a_{2} + a_{6} = a_{1} + d + a_{1} + 5 d = 2 a_{1} + 6 d ,，故A正确；B.$\begin{array}{r}{2 b_{4} = - 2 a_{1} - 1 0 d , b_{2} + b_{6} = a_{1} + 2 d - 3 a_{1} - 2 4 d = - 2 a_{1} - 2 2 d ,} \end{array}$

若$2 b_{4} = b_{2} + b_{6}$，则$- 2 a_{1} - 1 0 d = - 2 a_{1} - 2 2 d$，即d＝0，不合题意，故B错误；

$a_{4}^{2} = a_{2} a_{8}$

$(a_{1} + 3 d)^{2} = (a_{1} + d) (a_{1} + 7 d)$

即$a_{1}{}^{2} + f a_{1} d + g d^{2} = a_{1}{}^{2} + 8 a_{1} d + 7 d^{2}$，得$a_{1} d = d^{2} .$

$\because d \neq 0, \therefore a_{1} = d,$

$\frac{\mathrm{a}_{1}}{\mathrm{d}} \leqslant 1$

${}_{4}^{2} = b_{2} b_{8}$

$\left(- \mathrm{a}_{1} - 5 \mathrm{d}\right)^{2} = \left(\mathrm{a}_{1} + 2 \mathrm{d}\right) \left(- 5 \mathrm{a}_{1} - 5 5 \mathrm{d}\right)$

$2 \left(\frac{a_{1}}{d}\right)^{2} + 2 5 \frac{a_{1}}{d} + 4 5 = 0$

$\frac{\mathrm{a}_{1}}{\mathrm{d}}$

$\frac{a_{1}}{d} \leqslant 1$