依题意，数列\( \left\{{a}_{n}\right\}\)是公差为\(d\)的等差数列，数列\( \left\{{a}_{1}{a}_{n}\right\}\)为递减数列，

所以\( {a}_{1}{a}_{n}>{a}_{1}{a}_{n+1}\)，\( {a}_{1}{a}_{n}>{a}_{1}\left({a}_{n}+d\right)\)，\( {a}_{1}{a}_{n}>{a}_{1}{a}_{n}+{a}_{1}d,{a}_{1}d<0\).

故选：D

在公差不为零的等差数列\( \left\{{a}_{n}\right\}\)中，若\( 3{a}_{m}={a}_{1}+{a}_{2}+{a}_{3}\)，则\( m=\)（       ）

∵\( 2{a}_{2}={a}_{1}+{a}_{3}\)，则\( 3{a}_{m}={a}_{1}+{a}_{2}+{a}_{3}=3{a}_{2}\)

∴\( m=2\)

故选：B．

记数列\( \left\{{a}_{n}\right\}\)是等差数列，下列结论中不恒成立的是（       ）

设等差数列\( \left\{{a}_{n}\right\}\)的首项为\( {a}_{1}\)，公差为\( d\)，则

对于A，由数列\( \left\{{a}_{n}\right\}\)是等差数列及\( {a}_{1}+{a}_{2}>0\)，所以可取\( {a}_{1}=1，{a}_{2}=0，{a}_{3}=-1\)，所以\( {a}_{2}+{a}_{3}>0\)不成立，故A正确；

对于B，由数列\( \left\{{a}_{n}\right\}\)是等差数列，所以\( 2{a}_{2}={a}_{1}+{a}_{3}<0\)，所以\( {a}_{2}<0\)恒成立，故B不正确；

对于C, 由数列\( \left\{{a}_{n}\right\}\)是等差数列，\( {a}_{1}<{a}_{2}\)可取\( {a}_{1}=-3，{a}_{2}=-2，{a}_{3}=-1\)，所以\( {a}_{2}>\sqrt{{a}_{1}{a}_{3}}\)不成立，故C正确；

对于D，由数列\( \left\{{a}_{n}\right\}\)是等差数列，得\( \left({a}_{2}-{a}_{1}\right)\left({a}_{2}-{a}_{3}\right)=-{d}^{2}\le 0\)，无论\( {a}_{1}\)为何值，均有\( \left({a}_{2}-{a}_{1}\right)\left({a}_{2}-{a}_{3}\right)\le 0\)所以若\( {a}_{1}<0\)，则\( \left({a}_{2}-{a}_{1}\right)\left({a}_{2}-{a}_{3}\right)>0\)恒不成立，故D正确.

故选：ACD

已知等差数列\( \left\{{a}_{n}\right\}\)的公差是\(d\)，且\( {a}_{8}+{a}_{9}+{a}_{10}=36\)，则\( {a}_{1}d\)的最大值为___．

因为\( {a}_{8}+{a}_{9}+{a}_{10}=36\)，

所以\( 3{a}_{1}+24d=36\)，

\( {a}_{1}+8d=12\)，

所以\( {a}_{1}d=(12-8d)d=-8{d}^{2}+12d=-8{\left(d-\frac{3}{4}\right)}^{2}+\frac{9}{2}\)，

所以当\( d=\frac{3}{4}\)时，\( {a}_{1}d\)取最大值，最大值为\( \frac{9}{2}\).

故答案为：\( \frac{9}{2}\)

若数列\( \left\{2{a}_{n}+1\right\}\)是等差数列，其公差\( d=1\)，且\({a_3} = 5\)，则\( {a}_{10}=\)___．

∵数列\( \left\{2{a}_{n}+1\right\}\)是等差数列，则\( \left(2{a}_{10}+1\right)=\left(2{a}_{3}+1\right)+7d=18\)

∴\( {a}_{10}=\frac{17}{2}\)

故答案为：\( \frac{17}{2}\)．

求证：\( \left\{\frac{1}{{a}_{n}}\right\}\)是等差数列，并求\( \left\{{a}_{n}\right\}\)的通项公式；

若数列\( \left\{{a}_{n}\right\}\)满足：\( {a}_{1}=\frac{1}{2}\)，\( {a}_{n+1}=\frac{{a}_{n}}{2{a}_{n}+1}\)则第三项\( {a}_{3}=\)___，它的通项公式\({a}_{n}=\)___．

由\( {a}_{n+1}=\frac{{a}_{n}}{2{a}_{n}+1}\)可得\( \frac{1}{{a}_{n+1}}=\frac{2{a}_{n}+1}{{a}_{n}}=\frac{1}{{a}_{n}}+2\)，

所以\( \left\{\frac{1}{{a}_{n}}\right\}\)为首项是\( \frac{1}{{a}_{1}}=2\)公差为\( 2\)的等差数列，

所以\( \frac{1}{{a}_{n}}=2+(n-1)\times 2=2n\)，

所以\( {a}_{n}=\frac{1}{2n}\)，

所以\( {a}_{3}=\frac{1}{6}\).

故答案为：\( \frac{1}{6},\frac{1}{2n}\)

设等差数列\(\left\{ {{a_n}} \right\}\)的前\(n\)项和为\({S_n}\)，且\({S_{2021}} > 0\)，\({S_{2022}} < 0\)，则当\({S_n}\)最大时，\(n = \)（       ）

由\({S_{2021}} > 0\)可得\({S_{2021}} = \frac{{2021({a_1} + {a_{2021}})}}{2} = 2021{a_{1011}} > 0\)，即\({a_{1011}} > 0\)，

由\({S_{2022}} < 0\)可得\({S_{2022}} = \frac{{2022({a_1} + {a_{2022}})}}{2} = 1011({a_{1011}} + {a_{1012}}) < 0\)，即\({a_{1011}} + {a_{1012}} < 0\)，

故\({a_{1011}} > 0,{a_{1012}} < 0\)，则数列\(\left\{ {{a_n}} \right\}\)的前1011项为正数，从第1012项为负数的递减数列，

故当\({S_n}\)最大时，\(n=1011\)，

故选：B

在等差数列\(\left\{ {{a_n}} \right\}\)中，其前n项和为\({S_n}\)，若\({a_3} + {a_4} + {a_5} + {a_6} = 45\)，则\({S_8} = \)（       ）

∵\(\left\{ {{a_n}} \right\}\)是等差数列，由等差数列性质得\({a_3} + {a_4} + {a_5} + {a_6} = 2\left( {{a_1} + {a_8}} \right) = 45\)，得\({a_1} + {a_8} = \frac{{45}}{2}\)，∴\({S_8} = \)\(\frac{{\left( {{a_1} + {a_8}} \right) \times 8}}{2} = \frac{{45}}{2} \times 4 = 90\).

故选：B.

已知等差数列\(\left\{ {{a_n}} \right\}\)满足\({a_1} + {a_3} + {a_5} = 12\)，\({a_{10}} + {a_{11}} + {a_{12}} = 24\)，则\(\left\{ {{a_n}} \right\}\)的前\(13\)项的和为（       ）

设等差数列\(\left\{ {{a_n}} \right\}\)公差为\(d\)，

\(\because {a_1} + {a_3} + {a_5} = 12\)，\({a_{10}} + {a_{11}} + {a_{12}} = 24\)，\(\therefore {a_{10}} + {a_{11}} + {a_{12}} - \left( {{a_1} + {a_3} + {a_5}} \right) = 24d = 12\)，

解得：\(d = \frac{1}{2}\)，\(\therefore {a_1} + {a_3} + {a_5} = 3{a_1} + 6d = 3{a_1} + 3 = 12\)，解得：\({a_1} = 3\)，

\(\therefore \left\{ {{a_n}} \right\}\)的前\(13\)项的和为\(13{a_1} + \frac{{13 \times 12}}{2}d = 39 + \frac{{13 \times 12}}{4} = 78\).

故选：C.

等差数列\(\left\{ {{a_n}} \right\}\)的前n项和为\({S_n}\)，则\({S_{19}} = \frac{{{a_1} + {a_{19}}}}{2} \times 19 = 19{a_{10}} > 0\)，有\({a_{10}} > 0\)，

\({S_{20}} = \frac{{{a_1} + {a_{20}}}}{2} \times 20 = 10({a_{10}} + {a_{11}}) < 0\)，有\({a_{11}} < - {a_{10}} < 0\)，显然数列\(\left\{ {{a_n}} \right\}\)是递减的，且\({a_{10}} \cdot {a_{11}} < 0\)，

因\({a_m} \cdot {a_{m + 1}} < 0\)，所以\(m = 10\).

故选：B

已知项数为\(n\)的等差数列\(\left\{ {{a_n}} \right\}\)的前\(6\)项和为\(10\)，最后\(6\)项和为\(110\)，所有项和为\(360\)，则\(n = \)（       ）

由题意知\({a_1} + {a_2} + \cdot \cdot \cdot + {a_6} = 10\)，\({a_n} + {a_{n - 1}} + \cdot \cdot \cdot + {a_{n - 5}} = 110\)，

两式相加得\(6\left( {{a_1} + {a_n}} \right) = 120\)，所以\({a_1} + {a_n} = 20\)，又\(\frac{{n\left( {{a_1} + {a_n}} \right)}}{2} = 360\)，所以\(n = 36\)．

故选：B．

已知等差数列\(\left\{ {{a_n}} \right\}\)的前\(n\)项和为\({S_n}\)，公差为\(d\)，若\({S_{10}} < {S_9} < {S_{11}}\)，则（       ）

因为\({S_{10}} < {S_9}\)，

\({S_{10}} < {S_{11}}\)，

所以\({S_{10}} - {S_9} = {a_{10}} < 0\)，

\({S_{11}} - {S_{10}} = {a_{11}} > 0\)，

故等差数列首项为负，公差为正，所以\(d > 0\)，\({a_1} < 0\)，

故A正确，B错误；由\({S_9} < {S_{11}}\)，

可知\({S_{11}} - {S_9} = {a_{10}} + {a_{11}} > 0\)，

所以\({S_{20}} = 10\left( {{a_1} + {a_{20}}} \right) = 10\left( {{a_{10}} + {a_{11}}} \right) > 0\)，

故C错误；

因为\({a_{11}} > 0\)，所以\({S_{21}} = 21{a_{11}} > 0\)，故D正确．

故选：AD

已知等差数列\(\left\{ {{a_n}} \right\}\)的前n项和为\({S_n}\)，若\({S_{12}} = 3{a_{11}}\)，则\({a_5} = \)___，\({S}_{9}=\)___．

等差数列\(\left\{ {{a_n}} \right\}\)中，\({S_{12}} = 12{a_1} + 66d = 3{a_{11}} = 3{a_1} + 30d\)，

所以\(12{a_1} + 66d = 3{a_1} + 30d\)，

即\({a_1} = - 4d\)，

所以\({a_5} = {a_1} + 4d = 0\)，\({S_9} = 9{a_5} = 0\)

故答案为：①0；②0.

等差数列\(\left\{ {{a_n}} \right\}\)的前\(n\)项和为\({S_n}\)，已知\({S_{10}} = 0\)，\({S_{15}} = 25\)，则\(n + {S_n}\)的最小值为___．

由\({S_n} = n{a_1} + \frac{{n\left( {n - 1} \right)d}}{2}\)，\({S_{10}} = 0\)，\({S_{15}} = 25\)得\(\left\{ {\begin{array}{*{20}{l}}
{10{a_1} + 45d = 0} \\\
{15{a_1} + 105d = 25}
\end{array}} \right.\)，

解得：\(\left\{ {\begin{array}{*{20}{l}}
{{a_1} = - 3} \\\
{d = \frac{2}{3}}
\end{array}} \right.\)，

则\({S_n} = - 3n + \frac{{n\left( {n - 1} \right)}}{2} \cdot \frac{2}{3}{\rm{ = }}\frac{1}{3}\left( {{n^2} - 10n} \right)\)．故\(n + {S_n} = \frac{1}{3}\left( {{n^2} - 7n} \right) = \frac{1}{3}\left[ {{{\left( {n - \frac{7}{2}} \right)}^2} - \frac{{49}}{4}} \right]\)．

由于\(n \in {{\rm{N}}^ * }\)，故当\(n = 3\)或4时，\({\left( {n + {S_n}} \right)_{\min }} = - 4\).

故答案为：\( - 4\)

求通项\({a_n}\)；

因为\(\left\{ {{a_n}} \right\}\)是等差数列，\({a_3} + 12{a_7} = {S_{12}}\)，所以\({a_1} + 2d + 12\left( {{a_1} + 6d} \right) = 12{a_1} + 66d\)，

即\({a_1} + 8d = 0\)，亦即\({a_9} = 0\)，

所以\({S_{17}} = \frac{{\left( {{a_1} + {a_{17}}} \right) \times 17}}{2} = 0,{S_{11}} - {S_6} = {a_7} + {a_8} + {a_9} + {a_{10}} + {a_{11}} = 5{a_9} = 0\)，\({a_8} \cdot {a_{10}} = \left( {{a_9} - d} \right) \cdot \left( {{a_9} + d} \right) = a_9^2 - {d^2} = - {d^2} \leqslant 0\)．

故选：ABC．

因为\({S_n} = \frac{{n\left( {{a_1} + {a_n}} \right)}}{2} = \frac{{n\left( {{a_5} + {a_{n - 4}}} \right)}}{2}\)，

又\({a_5} = 7,{a_{n - 4}} = 29,{S_n} = 198\)，

所以\(18n = 198\)，

所以\(n = 11\)，

故选：B．