高考数学等差题目及答案

高考数学等差题目及答案

一、单项选择题1.已知等差数列\(\{a_{n}\}\)中，\(a_{3}=5\)，\(a_{5}=9\)，则\(a_{7}\)等于（）A.12B.13C.14D.15答案：B2.等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，若\(a_{1}=2\)，\(S_{3}=12\)，则\(a_{6}\)等于（）A.8B.10C.12D.14答案：C3.设等差数列\(\{a_{n}\}\)的公差为\(d\)，若数列\(\{2^{a_{1}a_{n}}\}\)为递减数列，则（）A.\(d\lt0\)B.\(d\gt0\)C.\(a_{1}d\lt0\)D.\(a_{1}d\gt0\)答案：C4.已知等差数列\(\{a_{n}\}\)满足\(a_{1}+a_{2}+a_{3}+\cdots+a_{101}=0\)，则有（）A.\(a_{1}+a_{101}\gt0\)B.\(a_{2}+a_{100}\lt0\)C.\(a_{3}+a_{99}=0\)D.\(a_{51}=51\)答案：C5.等差数列\(\{a_{n}\}\)中，\(a_{1}+a_{4}+a_{7}=39\)，\(a_{3}+a_{6}+a_{9}=27\)，则数列\(\{a_{n}\}\)的前\(9\)项和\(S_{9}\)等于（）A.99B.66C.144D.297答案：A6.已知等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，若\(S_{11}=22\)，则\(a_{3}+a_{7}+a_{8}\)等于（）A.18B.12C.9D.6答案：D7.等差数列\(\{a_{n}\}\)中，\(a_{n-4}=30\)，且前\(9\)项的和\(S_{9}=18\)，前\(n\)项和\(S_{n}=240\)，则\(n\)的值为（）A.15B.16C.17D.18答案：A8.设等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，若\(a_{2}+a_{8}=15-a_{5}\)，则\(S_{9}\)等于（）A.18B.36C.45D.60答案：C9.已知等差数列\(\{a_{n}\}\)的公差\(d\neq0\)，且\(a_{1}\)，\(a_{3}\)，\(a_{13}\)成等比数列，若\(a_{1}=1\)，\(S_{n}\)为数列\(\{a_{n}\}\)的前\(n\)项和，则\(\frac{2S_{n}+16}{a_{n}+3}\)的最小值为（）A.4B.3C.\(2\sqrt{3}-2\)D.\(\frac{9}{2}\)答案：A10.已知数列\(\{a_{n}\}\)满足\(a_{n+1}=a_{n}-a_{n-1}(n\geq2)\)，\(a_{1}=1\)，\(a_{2}=3\)，记\(S_{n}=a_{1}+a_{2}+\cdots+a_{n}\)，则下列结论正确的是（）A.\(a_{100}=-1\)，\(S_{100}=5\)B.\(a_{100}=-3\)，\(S_{100}=5\)C.\(a_{100}=-3\)，\(S_{100}=2\)D.\(a_{100}=-1\)，\(S_{100}=2\)答案：A二、多项选择题1.已知等差数列\(\{a_{n}\}\)的公差\(d\neq0\)，前\(n\)项和为\(S_{n}\)，若\(S_{6}=S_{12}\)，则下列结论中正确的有（）A.\(a_{1}:d=-17:2\)B.\(S_{18}=0\)C.当\(d\gt0\)时，\(a_{6}+a_{14}\gt0\)D.当\(d\lt0\)时，\(\verta_{6}\vert\gt\verta_{14}\vert\)答案：ABC2.设等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，公差为\(d\)，已知\(a_{3}=12\)，\(S_{12}\gt0\)，\(S_{13}\lt0\)，则下列结论正确的有（）A.\(a_{6}+a_{7}\gt0\)B.\(a_{7}\lt0\)C.\(d\lt0\)D.\(S_{5}=60\)答案：ABCD3.已知数列\(\{a_{n}\}\)是等差数列，\(S_{n}\)是其前\(n\)项和，若\(a_{1}+a_{2}^{2}=-3\)，\(S_{5}=10\)，则下列说法正确的有（）A.\(a_{2}=-1\)B.\(a_{1}=-4\)C.\(S_{10}=45\)D.\(d=2\)答案：ABD4.等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，若\(a_{4}\)，\(a_{10}\)是方程\(x^{2}-8x+1=0\)的两根，则（）A.\(a_{7}=4\)B.\(S_{13}=52\)C.\(a_{7}=\pm4\)D.\(S_{11}=66\)答案：AB5.已知等差数列\(\{a_{n}\}\)中，\(a_{1}=-1\)，公差\(d=2\)，则（）A.数列\(\{a_{n}\}\)是递增数列B.通项公式\(a_{n}=2n-3\)C.数列\(\{a_{n}\}\)是递减数列D.前\(n\)项和\(S_{n}=n^{2}-2n\)答案：ABD6.等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，若\(S_{10}=10\)，\(S_{20}=30\)，则（）A.\(S_{30}=60\)B.\(S_{30}=70\)C.\(a_{11}+a_{12}+\cdots+a_{20}=20\)D.\(a_{21}+a_{22}+\cdots+a_{30}=40\)答案：BC7.设\(\{a_{n}\}\)是等差数列，\(S_{n}\)为其前\(n\)项和，且\(S_{5}\ltS_{6}\)，\(S_{6}=S_{7}\gtS_{8}\)，则下列结论正确的有（）A.\(d\lt0\)B.\(a_{7}=0\)C.\(S_{9}\gtS_{5}\)D.\(S_{6}\)与\(S_{7}\)均为\(S_{n}\)的最大值答案：ABD8.已知等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，\(a_{5}=5\)，\(S_{5}=15\)，则数列\(\{\frac{1}{a_{n}a_{n+1}}\}\)的前\(100\)项和为（）A.\(\frac{100}{101}\)B.\(\frac{99}{101}\)C.利用裂项相消法求和D.利用错位相减法求和答案：AC9.已知等差数列\(\{a_{n}\}\)的首项\(a_{1}=1\)，公差\(d=2\)，其前\(n\)项和为\(S_{n}\)，则（）A.\(a_{n}=2n-1\)B.\(S_{n}=n^{2}\)C.数列\(\{\frac{1}{S_{n}}\}\)的前\(n\)项和为\(\frac{2n}{n+1}\)D.数列\(\{(-1)^{n}a_{n}\}\)的前\(20\)项和为\(20\)答案：ABD10.等差数列\(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{n}=-512\)，\(S_{n}=-1022\)，则公差\(d\)的值为（）A.-171B.-172C.-173D.-174答案：无正确选项（注：本题经计算公差\(d=-171\)，应选A，但原选项可能存在错误）三、判断题1.若一个数列从第二项起每一项与它的前一项的差都是常数，那么这个数列是等差数列。（×）2.等差数列\(\{a_{n}\}\)中，若\(a_{m}+a_{n}=a_{p}+a_{q}\)，则\(m+n=p+q\)。（×）3.等差数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}\)，则\(S_{n}\)，\(S_{2n}-S_{n}\)，\(S_{3n}-S_{2n}\)仍成等差数列。（√）4.若数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}=An^{2}+Bn+C\)（\(A\)，\(B\)，\(C\)为常数），则数列\(\{a_{n}\}\)一定是等差数列。（×）5.等差数列\(\{a_{n}\}\)中，若公差\(d\gt0\)，则数列\(\{a_{n}\}\)是递增数列。（√）6.已知等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，若\(S_{10}=S_{20}\)，则\(S_{30}=0\)。（√）7.若\(\{a_{n}\}\)是等差数列，\(a_{1}\gt0\)，\(d\lt0\)，且\(a_{3}+a_{9}=0\)，则\(\{a_{n}\}\)的前\(9\)项和最大。（√）8.等差数列\(\{a_{n}\}\)中，\(a_{5}=a_{3}+2d\)。（√）9.若\(\{a_{n}\}\)，\(\{b_{n}\}\)都是等差数列，则\(\{a_{n}+b_{n}\}\)也是等差数列。（√）10.等差数列\(\{a_{n}\}\)的通项公式\(a_{n}=a_{1}+(n-1)d\)可以变形为\(a_{n}=dn+(a_{1}-d)\)，从函数角度看，\(a_{n}\)是关于\(n\)的一次函数。（×）四、简答题1.已知等差数列\(\{a_{n}\}\)中，\(a_{3}=7\)，\(a_{5}+a_{7}=32\)，求\(\{a_{n}\}\)的通项公式。答案：设等差数列\(\{a_{n}\}\)的公差为\(d\)。因为\(a_{3}=7\)，所以\(a_{1}+2d=7\)。又\(a_{5}+a_{7}=32\)，即\((a_{1}+4d)+(a_{1}+6d)=32\)，\(2a_{1}+10d=32\)，\(a_{1}+5d=16\)。联立\(\begin{cases}a_{1}+2d=7\\a_{1}+5d=16\end{cases}\)，两式相减得\(3d=9\)，\(d=3\)，代入\(a_{1}+2d=7\)得\(a_{1}=1\)。所以\(a_{n}=a_{1}+(n-1)d=1+3(n-1)=3n-2\)。2.等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，已知\(a_{1}=10\)，\(a_{2}\)为整数，且\(S_{n}\leqS_{4}\)。求\(\{a_{n}\}\)的通项公式。答案：因为\(S_{n}\leqS_{4}\)，所以\(a_{4}\geq0\)，\(a_{5}\leq0\)。设公差为\(d\)，\(a_{1}=10\)，则\(a_{4}=10+3d\geq0\)，\(a_{5}=10+4d\leq0\)，解得\(-\frac{10}{3}\leqd\leq-\frac{5}{2}\)。又\(a_{2}=10+d\)为整数，所以\(d=-3\)。那么\(a_{n}=a_{1}+(n-1)d=10-3(n-1)=13-3n\)。3.已知等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，\(a_{3}=5\)，\(S_{3}=9\)，求数列\(\{\frac{1}{a_{n}a_{n+1}}\}\)的前\(n\)项和\(T_{n}\)。答案：设等差数列\(\{a_{n}\}\)公差为\(d\)，由\(a_{3}=5\)，\(S_{3}=9\)可得\(\begin{cases}a_{1}+2d=5\\3a_{1}+\frac{3\times2}{2}d