4.2.1 等差数列的概念2(性质运用)
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基础知识
等差数列基本性质
若数列\(\{a_n\}\)是首项为\(a_1\),公差为\(d\)的等差数列,其中\(m\) ,\(n\) ,\(s\),\(t∈N^*\),
它具有以下性质:
(1) \(a_n=a_m+(n-m) d\);
证明 由等差数列通项公式可得\(a_n=a_1+(n-1)d\),\(a_m=a_1+(m-1)d\),
\(\qquad\) 两式相减可得\(a_n-a_m=(n-m) d\),即\(a_n=a_m+(n-m) d\).
意义 求等差数列任一项\(a_k\)或通项公式\(a_n\),不一定要求\(a_1\),可利用任一项(非\(a_k\)即可).
例 若等差数列\(\{a_n\}\)中,\(a_3=4\),\(d=2\),则\(a_6=\)\(\underline{\quad \quad}\) .
解 \(a_6=a_3+3d=10\).
(2) \(d=\dfrac{a_n-a_m}{n-m}\);
证明 由性质\(a_n=a_m+(n-m) d\)可得 \(d=\dfrac{a_n-a_m}{n-m}\).
意义 利用等差数列任意两项可求公差.
例 若等差数列\(\{a_n\}\)中,\(a_3=4\),\(a_6=10\),则公差\(d=\) \(\underline{\quad \quad}\).
解 \(d=\dfrac{a_6-a_3}{6-3}=\dfrac{6}{3}=2\).
(3)若 $m+n=s+t $, 则 \(a_m+a_n=a_s+a_t\);
证明 由等差数列通项公式可得
\(\qquad\) \(a_m+a_n=a_1+(m-1)d+a_1+(n-1)d=2a_1+(m+n-2)d\),
\(\qquad\) \(a_s+a_t=a_1+(s-1)d+a_1+(t-1)d=2a_1+(s+t-2)d\),
\(\qquad\) \(\because m+n=s+t\),\(\therefore 2a_1+(m+n-2)d=2a_1+(s+t-2)d\)
\(\qquad\) 即\(a_m+a_n=a_s+a_t\).
意义 下标和相等,其对应项的和相等.
例 \(a_2+a_8=a_1+a_9=2a_5\),但\(a_2+a_8\)不一定等于\(a_{10}\).
(4)下标成等差数列且公差为\(m\)的项\(a_k\) ,\(a_{k+m}\) ,\(a_{k+2m}\) ,\(…(k ,m∈N^*)\)组成公差为\(md\)的等差数列;
证明 \(a_{k+n m}-a_{k+(n-1) m}=a_1+(k+n m-1) d-a_1-(k+n m-m-1) d=m d\),得证.
例 若\(\{a_n\}\)是等差数列,则\(a_3\),\(a_6\),\(a_9\)是公差为\(3d\)的等差数列, \(\{a_{2n-1}\}\)均是公差为\(2d\)的等差数列.
(5) 数列\(\{λa_n+b\}\)\((λ、b\)是常数\()\)是公差是\(λb\)的等差数列;
证明 利用等差数列的定义可证.
(6) 若数列\(\{b_n \}\)也是等差数列,则数列\(\{ka_n±mb_n+b\}\)\((k,m,b\)为非零常数\()\)也是等差数列;
证明 利用等差数列的定义可证.
基本方法
【题型1】 等差数列性质的应用
【典题1】 设\(\{a_n \}\)为等差数列,若\(a_3+a_4+a_5+a_6+a_7=450\),求\(a_2+a_8\).
解析 方法1 \(\because \{a_n \}\)为等差数列,设首项为\(a_1\),公差为\(d\),
\(\therefore a_3+a_4+⋯+a_7=a_1+2d+a_1+3d+⋯+a_1+6d=5a_1+20d\),
即\(5a_1+20d=450\),
\(\therefore a_1+4d=90\).
\(\therefore a_2+a_8=a_1+d+a_1+7d=2a_1+8d=180\).
方法2 \(\because a_3+a_7=a_4+a_6=2a_5=a_2+a_8\),
\(\therefore a_3+a_4+a_5+a_6+a_7=5a_5=450\),
\(\therefore a_5=90\),\(\therefore a_2+a_8=2a_5=180\).
点拨 方法1是利用通项公式\(a_n=a_1+(n-1)d\)处理,方法2注意到各项中下标的特点,
利用等差数列的性质处理,这样来得更简便.
【典题2】(多选)设\(d\)为正项等差数列\(\{a_n\}\)的公差,若\(d>0\),\(a_3=2\),则( )
A.\(a_2 a_4<4\) \(\qquad \qquad\) B. \(a_2^2+a_4 \geq \dfrac{15}{4}\) \(\qquad \qquad\) C. \(\dfrac{1}{a_1}+\dfrac{1}{a_5}>1\) \(\qquad \qquad\) D.\(a_1 a_5>a_2 a_4\)
解析 由题设知: \(\left\{\begin{array}{l}
d>0 \\
a_1=2-2 d>0
\end{array}\right.\),解得\(0<d<1\),
\(\because a_2 a_4=(2-d)(2+d)=4-d^2<4\),\(\therefore A\)正确;
\(\because a_2^2+a_4=(2-d)^2+2+d=d^2-3d+6>4>\dfrac{15}{4}\),
\(\therefore B\)正确;
\(\because a_1+a_5=2 a_3=4, \dfrac{1}{a_1}+\dfrac{1}{a_5}=\dfrac{1}{4}\left(\dfrac{1}{a_1}+\dfrac{1}{a_5}\right)\left(a_1+a_5\right)\)
\(=\dfrac{1}{4}\left(2+\dfrac{a_5}{a_1}+\dfrac{a_1}{a_5}\right)>\dfrac{1}{4}\left(2+2 \sqrt{\dfrac{a_5}{a_1} \cdot \dfrac{a_1}{a_5}}\right)=1\),
\(\therefore C\)正确;
\(\because a_1 a_5-a_2 a_4=(2-2d)(2+2d)-(2-d)(2+d)=-3d^2<0\),
\(\therefore a_1 a_5<a_2 a_4\),\(D\)错误.
故选:\(ABC\).
【巩固练习】
1.在等差数列\(\{a_n\}\)中,\(a_3=7\),\(a_5+a_7=32\),则\(\{a_n\}\)的公差\(d=\)( )
A.\(5\) \(\qquad \qquad \qquad \qquad\) B.\(4\) \(\qquad \qquad \qquad \qquad\) C.\(3\) \(\qquad \qquad \qquad \qquad\) D.\(2\)
2.已知等差数列\(\{a_n\}\)满足\(a_1+a_3+a_5=18\),\(a_3+a_5+a_7=30\),则\(a_2+a_4+a_6=\)( )
A.\(20\) \(\qquad \qquad \qquad \qquad\) B.\(24\) \(\qquad \qquad \qquad \qquad\) C.\(26\) \(\qquad \qquad \qquad \qquad\) D.\(28\)
3.在等差数列\(\{a_n\}\)中,若 \(a_1-a_4-a_8-a_{12}+a_{15}=\dfrac{\pi}{4}\),则\(\sin(a_3+a_{13} )\)的值为( )
A.\(\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) B.\(1\) \(\qquad \qquad \qquad \qquad\) C.\(-1\) \(\qquad \qquad \qquad \qquad\) D.\(0\)
4.等差数列\(\{a_n\}\)中,\(a_2+a_5+a_8=12\),那么关于\(x\)的方程\(x^2+(a_4+a_6)x+10=0\)( )
A.无实根 \(\qquad \qquad\) B.有两个相等实根 \(\qquad \qquad\) C.有两个不等实根\(\qquad \qquad\) D.不能确定有无实根
参考答案
-
答案 \(C\)
解析 等差数列\(\{a_n\}\)中,\(a_5+a_7=2a_6=32\),所以\(a_6=16\),
因为\(a_3=7\),则 \(d=\dfrac{a_6-a_3}{6-3}=3\).
故选:\(C\). -
答案 \(B\)
解析 \(\because\) 等差数列\(\{a_n\}\)满足\(a_1+a_3+a_5=18\),\(a_3+a_5+a_7=30\),
设公差为\(d\),
相减可得\(6d=30-18=12\),\(\therefore d=2\).
则\(a_2+a_4+a_6=a_1+a_3+a_5+3d=24\),
故选:\(B\). -
答案 \(C\)
解析 由等差数列的性质可知,\(a_1+a_{15}=a_4+a_{12}\),
则 \(a_1-a_4-a_8-a_{12}+a_{15}=-a_8=\dfrac{\pi}{4}\),解得 \(a_8=-\dfrac{\pi}{4}\),
\(\because a_3+a_{13}=2a_8\),
\(\therefore \sin \left(a_3+a_{13}\right)=\sin \left(2 a_8\right)=\sin \left(-\dfrac{\pi}{2}\right)=-1\).
故选:\(C\). -
答案 \(C\)
解析 等差数列\(\{a_n\}\)中,\(a_2+a_5+a_8=12=3a_5\),解得\(a_5=4\).
那么关于\(x\)的方程\(x^2+(a_4+a_6 )x+10=0\)
化为:\(x^2+8x+10=0\),\(△=8^2-4×10>0\).
\(\therefore\) 方程有两个不等实数根.
故选:\(C\).
【题型2】 等差数列的综合问题
【典题1】 已知三个数成等差数列,其和为\(15\),首末两项的积为\(9\),求这三个数.
解析 由题意,可设这三个数分别为\(a-d\),\(a\),\(a+d\),
则 \(\left\{\begin{array}{l}
(a-d)+a+(a+d)=15 \\
(a-d)(a+d)=9
\end{array}\right.\),解得 \(\left\{\begin{array}{l}
a=5 \\
d=4
\end{array}\right.\)或 \(\left\{\begin{array}{l}
a=5 \\
d=-4
\end{array}\right.\),
所以,当\(d=4\)时,这三个数为\(1,5,9\);
当\(d=-4\)时,这三个数为\(9,5,1\).
所以这三个数为\(1,5,9\)或\(9,5,1\).
点拨 设三个数分别为\(a-d\),\(a\),\(a+d\)比较好,计算简便些.
【典题2】已知数列\(\{a_n\}\)的前\(n\)项和是\(S_n\),且对任意\(n∈N^*\),有\(\dfrac{S_n}{n}\)是\(1\)和\(a_n\)的等差中项.
(1)求证:\(a_3-2a_2+1=0\);
(2)若\(a_4=7\),求数列\(\{a_n\}\)的通项公式.
解析 (1)证明:\(\because\)数列\(\{a_n\}\)的前\(n\)项和是\(S_n\),且对任意\(n∈N^*\),有\(\dfrac{S_n}{n}\)是\(1\)和\(a_n\)的等差中项.
\(\therefore 2 \times \dfrac{S_1}{1}=2 a_1=1+a_1\),解得\(a_1=1\),
\(2 \times \dfrac{S_3}{3}=2 \times \dfrac{1+a_2+a_3}{3}=1+a_3\),
整理得\(a_3-2a_2+1=0\).
(2)\(\because\) \(\dfrac{S_n}{n}\)是\(1\)和\(a_n\)的等差中项, \(\therefore 2 \times \dfrac{s_n}{n}=1+a_n\),\(\therefore 2S_n=n+na_n\),
\(2S_{n-1}=n-1+(n-1)a_{n-1}\),
两式相减可得\((n-2)a_n-(n-1)a_{n-1}+1=0\),
\(\therefore (n-1)a_{n+1}-na_n+1=0\),
整理可得,\(2a_n=a_{n-1}+a_{n+1} (n≥2)\),
故数列\(\{a_n\}\)是等差数列.
又\(a_4=7\),\(a_1=1\),\(\therefore\) 公差 \(d=\dfrac{a_4-a_1}{4-1}=2\),
\(\therefore a_n=a_1+(n-1)d=1+2(n-1)=2n-1\).
【巩固练习】
1.设数列\(\{a_n\}\)是等差数列,\(a_p=q\),\(a_q=p(p≠q)\),试求 \(a_{p+q}\).
2.一个等差数列的首项为\(\dfrac{1}{25}\),公差\(d>0\),从第\(10\)项起每一项都大于\(1\),求公差\(d\)的范围.
3.已知数列\(\{a_n\}\)满足\(a_1=2a\), \(a_n=2 a-\dfrac{a^2}{a_n-1}(n \geq 2)\),其中\(a\)是不为\(0\)的常数,令 \(b_n=\dfrac{1}{a_n-a}\).
(1)求证:数列\(\{b_n\}\)是等差数列;
(2)求数列\(\{a_n\}\)的通项公式.
参考答案
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答案 \(0\)
解析 设数列\(\{a_n \}\)的公差为\(d\),
\(\because a_p=a_q+(p-q)d\), \(\therefore d=\dfrac{a_p-a_q}{p-q}=\dfrac{q-p}{p-q}=-1\).
从而 \(a_{p+q}=a_p+q d=q+q \times(-1)=0\),
\(\therefore a_{p+q}=0\). -
答案 \(\left(\dfrac{8}{75}, \dfrac{3}{25}\right]\)
解析 设等差数列为\(\{a_n \}\),
由\(d>0\),知\(a_1<a_2<⋯<a_9<a_{10}<a_{11}…\),
依题意,有 \(\left\{\begin{array}{l} 1<a_{10}<a_{11}<\cdots \\ a_1<a_2<\cdots<a_9 \leq 1 \end{array}\right.\),
即\(\left\{\begin{array} { l } { a _ { 1 0 } > 1 } \\ { a _ { 9 } \leq 1 } \end{array} \Leftrightarrow \left\{\begin{array}{l} \dfrac{1}{25}+(10-1) d>1 \\ \dfrac{1}{25}+(9-1) d \leq 1 \end{array}\right.\right.\),
解得 \(\dfrac{8}{75}<d \leq \dfrac{3}{25}\),
即公差\(d\)的取值范围是\(\left(\dfrac{8}{75}, \dfrac{3}{25}\right]\). -
答案 \(a_n=a\left(1+\dfrac{1}{n}\right)\)
解析 (1) \(\because a_n=2 a-\dfrac{a^2}{a_{n-1}}(n \geq 2)\),
\(\therefore b_n=\dfrac{1}{a_n-a}=\dfrac{1}{a-\dfrac{a^2}{a_{n-1}}}=\dfrac{a_{n-1}}{a\left(a_{n-1}-a\right)}(n \geq 2)\),
\(\therefore b_n-b_{n-1}=\dfrac{a_{n-1}}{a\left(a_{n-1}-a\right)}-\dfrac{1}{a_{n-1}-a}=\dfrac{1}{a}(n \geq 2)\),
\(\therefore\)数列\(\{b_n\}\)是公差为 \(\dfrac{1}{a}\)的等差数列.
(2) \(\because b_1=\dfrac{1}{a_1-a}=\dfrac{1}{a}\),
故由(1)得: \(b_n=\dfrac{1}{a}+(n-1) \times \dfrac{1}{a}=\dfrac{n}{a}\).
即 \(\dfrac{1}{a_n-a}=\dfrac{n}{a}\),得\(a_n=a\left(1+\dfrac{1}{n}\right)\).
分层练习
【A组---基础题】
1.在等差数列\(\{a_n\}\)中,\(a_7-a_3=2\),\(a_4=1\),则\(a_{12}=\) ( )
A.\(5\) \(\qquad \qquad \qquad \qquad\) B.\(4\) \(\qquad \qquad \qquad \qquad\) C.\(3\) \(\qquad \qquad \qquad \qquad\) D.\(2\)
2.在等差数列\(\{a_n\}\)中,若\(a_1+a_5=10\),\(a_4=7\),则( )
A.\(a_6=10\) \(\qquad \qquad\) B.\(a_n=2n-1\) \(\qquad \qquad\) C.\(a_n=n+3\) \(\qquad \qquad\) D.\(a_n=-2n+3\)
3.新广中上月开展植树活动以来,学校环境愈发美丽.尤其是黄花风铃木,金黄的花朵挂满枝头,好不烂漫,俨然成了师生的热门打卡景点.书院数学兴趣小组的同学们通过调查发现:我校的黄花风铃树主要分布在孔子行教像旁(\(A\)处)、一食堂旁(\(B\)处)、高二教学楼旁(\(C\)处),如果把\(C\)处的\(5\)株移到\(B\)处,则\(A\)、\(B\)、\(C\)三处的株数刚好构成等差数列,已知B处现有\(11\)株,那么这三处共有黄花风铃树( )
A.\(36\)株 \(\qquad \qquad \qquad \qquad\) B.\(41\)株 \(\qquad \qquad \qquad \qquad\) C.\(48\)株 \(\qquad \qquad \qquad \qquad\) D.\(51\)株
4.已知等差数列\(\{a_n\}\)中,\(a_2\),\(a_8\)是\(2x^2-16x-1=0\)的两根,则\((a_3+a_7 )^2-a_5=\) ( )
A.\(248\) \(\qquad \qquad \qquad \qquad\) B.\(60\) \(\qquad \qquad \qquad \qquad\) C.\(12\) \(\qquad \qquad \qquad \qquad\) D.\(4\)
5.已知数列\(\{a_n\}\),\(\{b_n\}\)都是等差数列,\(a_1=1\),\(b_1=5\),且\(a_{21}-b_{21}=34\),则\(a_{11}-b_{11}\)的值( )
A.\(-17\) \(\qquad \qquad \qquad \qquad\) B.\(-15\) \(\qquad \qquad \qquad \qquad\) C.\(17\) \(\qquad \qquad \qquad \qquad\) D.\(15\)
6.设\(\{a_n\}\)是等差数列,且\(a_1=3\),\(a_2+a_4=14\),若\(a_m=41\),则\(m=\)\(\underline{\quad \quad}\).
7.若数列\(\{a_n\}\)是等差数列,且\(a_1+a_8+a_15=π\),则\(\tan(a_4+a_{12})=\)\(\underline{\quad \quad}\).
8.已知数列\(\{a_n\}\)满足\(2a_n=a_{n-1}+a_{n+1} (n≥2)\),\(a_1+a_3+a_5=6\),\(a_2+a_4+a_6=18\),则\(a_3+a_4=\)\(\underline{\quad \quad}\).
9.在等差数列\(\{a_n \}\)中,已知\(a_5=10\),\(a_{12}>31\),求公差\(d\)的取值范围.
10.已知三个数成等差数列并且是递增数列,它们的和为\(18\),平方和为\(116\),求这三个数.
11.证明: \(2, \sqrt{5}, 7\)不可能是某个等差数列中的三项.
- 已知等差数列\(\{a_n\}\)满足\(a_2+a_3+a_4=18\),\(a_2 a_3 a_4=66\),求数列\(\{a_n\}\)的通项公式.
参考答案
-
答案 \(A\)
解析 等差数列\(\{a_n\}\)中,\(a_7-a_3=4d=2\),
因为\(a_4=1\),则\(a_{12}=a_4+8d=1+4=5\).
故选:\(A\). -
答案 \(B\)
解析 因为等差数列\(\{a_n\}\)中,\(a_1+a_5=2a_3=10\),\(a_4=7\),
所以\(a_3=5\),\(d=2\),
所以\(a_n=a_3+(n-3)d=5+2(n-3)=2n-1\),\(B\)正确,
\(a_5=9\),\(A\)错误.
故选:\(B\). -
答案 \(C\)
解析 设\(A\)、\(B\)、\(C\)三处的株数构成等差数列\(\{a_n\}\),
显然\(a_2=16\),故\(a_1+a_3=2a_2=32\),
故这三处共有黄花风铃树共\(48\)株.
故选:\(C\). -
答案 \(B\)
解析 \(\because\) 等差数列\(\{a_n\}\)中,\(a_2\),\(a_8\)是\(2x^2-16x-1=0\)的两根,
\(\therefore a_2+a_8=2a_5=8\),即\(a_5=4\),
\(\therefore (a_3+a_7 )^2-a_5=(a_2+a_8 )^2-a_5=64-4=60\).
故选:\(B\). -
答案 \(D\)
解析 因为数列\(\{a_n\}\),\(\{b_n\}\)都是等差数列,\(a_1=1\),\(b_1=5\),
因为\(a_{21}-b_{21}=34\),
则 \(2\left(a_{11}-b_{11}\right)=\left(a_1+a_{21}\right)-\left(b_1+b_{21}\right)=a_1-b_1+a_{21}-b_{21}=30\),
所以 \(a_{11}-b_{11}=15\).
故选:\(D\). -
答案 \(20\)
解析 设等差数列\(\{a_n\}\)的公差为\(d\),
\(\because a_2+a_4=14\),\(\therefore 2a_3=14\),解得\(a_3=7\),
\(\therefore 2d=a_3-a_1=7-3=4\),解得\(d=2\),
\(\therefore a_n=a_1+2(n-1)=3+2n-2=2n+1\),
\(\because a_m=41\),\(\therefore 2m+1=41\),解得\(m=20\).
故答案、为:\(20\). -
答案 \(-\sqrt{3}\)
解析 因为数列\(\{a_n\}\)是等差数列,由等差数列的性质可知\(a_1+a_8+a_{15}=3a_8=π\),
所以 \(a_8=\dfrac{\pi}{3}\),
则 \(\tan \left(a_4+a_{12}\right)=\tan \left(2 a_8\right)=\tan \dfrac{2 \pi}{3}=-\sqrt{3}\).
故答案为: \(-\sqrt{3}\). -
答案 \(8\)
解析 因为数列\(\{a_n\}\)满足\(2a_n=a_{n-1}+a_{n+1} (n≥2)\),
所以数列\(\{a_n\}\)为等差数列,
则\(a_1+a_3+a_5=6\),\(a_2+a_4+a_6=18\),
则\(a_3+a_4=2+6=8\). -
答案 \((3,+∞)\)
解析 由题意,可知 \(\left\{\begin{array}{l} a_1+4 d=10 \\ a_1+11 d>31 \end{array}\right.\),解得\(d>3\).
所以\(d\)的取值范围是\((3,+∞)\). -
答案 \(4,6,8\)
解析 解法一:设这三个数为\(a,b,c\),
则由题意,得 \(\left\{\begin{array}{l} 2 b=a+c \\ a+b+c=18 \\ a^2+b^2+c^2=116 \end{array}\right.\),
解得\(a=4\),\(b=6\),\(c=8\).
故这三个数是\(4,6,8\).
解法二:设这三个数为\(a-d\),\(a\),\(a+d\),
由已知,得 \(\left\{\begin{array}{l} (a-d)+a+(a+d)=18, (1) \\ (a-d)^2+a^2+(a+d)^2=116,(2) \end{array}\right.\),
由(1),得\(a=6\),代入(2),得\(d=±2\).
\(\because\) 该数列是递增的,\(\therefore d=-2\)舍去.
\(\therefore\)这三个数为\(4,6,8\). -
证明 假设 \(2, \sqrt{5}, 7\)不是同一个等差数列中的三项,分别设为\(a_m\),\(a_n\),\(a_p\),
则 \(d=\dfrac{7-2}{p-m}\)为有理数,
又 \(d=\dfrac{\sqrt{5}-2}{n-m}\)为无理数,矛盾.
所以,假设不成立,
即\(2, \sqrt{5}, 7\)不可能是同一个等差数列中的三项. -
答案 \(a_n=-5n+21\)或\(a_n=5n-9\)
解析 在等差数列\(\{a_n\}\)中,\(a_2+a_3+a_4=18\),
即\(3a_3=18\),所以\(a_3=6\),
则\(a_2+a_4=12\),\(a_2 a_4=11\),
解得 \(\left\{\begin{array}{l} a_2=11 \\ a_4=1 \end{array}\right.\)或 \(\left\{\begin{array}{l} a_2=1 \\ a_4=11 \end{array}\right.\),
当\(\left\{\begin{array}{l} a_2=11 \\ a_4=1 \end{array}\right.\)时,\(a_1=16\),\(d=-5\),所以\(a_n=-5n+21\);
当\(\left\{\begin{array}{l} a_2=1 \\ a_4=11 \end{array}\right.\)时,\(a_1=-4\),\(d=5\),所以\(a_n=5n-9\).
综上所述,数列\(\{a_n\}\)的通项公式为\(a_n=-5n+21\)或\(a_n=5n-9\).
【B组---提高题】
1.(多选)设正项等差数列\(\{a_n\}\)满足\((a_1+a_{10})^2=2a_2 a_9+20\),则( )
A.\(a_2 a_9\)的最大值为\(10\) \(\qquad \qquad \qquad \qquad \qquad\) B.\(a_2+a_9\)的最大值为 \(2 \sqrt{10}\)
C. \(\dfrac{1}{a_2^2}+\dfrac{1}{a_9^2}\)的最大值为\(\dfrac{1}{5}\) \(\qquad \qquad \qquad \qquad\) D.\(a_2^4+a_9^4\)的最小值为\(200\)
2.已知数列\(\{a_n\}\)为等差数列,公差\(d(d≠0)\),且满足\(a_3 a_4+2a_4 a_6+a_5 a_{12}=2024d\),则 \(\dfrac{1}{a_6}-\dfrac{1}{a_5}=\)\(\underline{\quad \quad}\).
参考答案
-
答案 \(ABD\)
解析 正项等差数列\(\{a_n\}\)满足\(\left(a_1+a_{10}\right)^2=2 a_2 a_9+20=\left(a_2+a_9\right)^2\),
所以\(a_2^2+a_9^2=20\).
① \(a_2 a_9 \leq \dfrac{1}{2}\left(a_2^2+a_9^2\right)=10\),
当且仅当 \(a_2=a_9=\sqrt{10}\)时成立,故\(A\)选项正确.
②由于\(\left(\dfrac{a_2+a_9}{2}\right)^2 \leq \dfrac{1}{2}\left(a_2^2+a_9^2\right)=10\),
所以\(\dfrac{a_2+a_9}{2} \leq \sqrt{10}\), \(a_2+a_9 \leq 2 \sqrt{10}\),
当且仅当\(a_2=a_9=\sqrt{10}\)时成立,故\(B\)选项正确.
③ \(\dfrac{1}{a_2^2}+\dfrac{1}{a_9^2}=\dfrac{a_2^2+a_9^2}{a_2^2 a_9^2}=\dfrac{20}{a_2^2 a_9^2} \geq \dfrac{20}{\left(\dfrac{a_2^2+a_9^2}{2}\right)^2}=\dfrac{20}{10^2}=\dfrac{1}{5}\),
当且仅当 \(a_2=a_9=\sqrt{10}\)时成立,
所以 \(\dfrac{1}{a_2^2}+\dfrac{1}{a_9^2}\)的最小值为\(\dfrac{1}{5}\),故\(C\)选项错误.
④结合①的结论,有 \(a_2^4+a_9^4=\left(a_2^2+a_9^2\right)^2-2 a_2^2 a_9^2=400-2 \times 10^2=200\),
当且仅当\(a_2=a_9=\sqrt{10}\)时成立,故\(D\)选项正确.
故选:\(ABD\). -
答案 \(-\dfrac{1}{506}\)
解析 由等差数列\(\{a_n\}\)满足\(a_3 a_4+2a_4 a_6+a_5 a_12=2024d\),
得\(a_4 (a_4-d)+2a_4 (a_4+2d)+(a_4+d)(a_4+8d)=2024d\),
即\(4a_4^2+12a_4 d+8d^2=2024d\),所以\(4(a_4+2d)(a_4+d)=2024d\),
即\(4a_6 a_5=2024d\),故\(a_6 a_5=506d\),
所以\(\dfrac{1}{a_6}-\dfrac{1}{a_5}=\dfrac{a_5-a_6}{a_6 a_5}=\dfrac{-d}{506 d}=-\dfrac{1}{506}\).
故答案为:\(-\dfrac{1}{506}\).
【C组---拓展题】
1.(2022•山西自主招生)对于公差为\(d(d≠0)\)的等差数列\(\{a_n\}\),求证:数列中不同两项之和仍是这一数列中的一项的充要条件是存在整数\(m≥-1\),使\(a_1=md\).
参考答案
- 证明 ①先证必要性:任取等差数列\(\{a_n\}\)中不同的两项\(a_s\),\(a_t\),\((s≠t)\),
存在整数\(k\)使得\(a_s+a_t=a_k\),
则\(2a_1+(s+t-2)d=a_1+(k-1)d\),得\(a_1=(k-s-t+1)d\),
故存在\(m\)使得\(m=k-s-t+1\),使得\(a_1=md\),\(m∈Z\),
再整\(m≥-1\):
反证法证明:假设当\(d≠0\)时,\(m≥-1\)不成立,则\(m<-1\)恒成立,
对于不同的两项\(a_1\),\(a_2\),应存在\(a_l\),使得\(a_1+a_2=a_l\),
即\((2m+1)d=md+(l-1)d\),
所以\(l=m+2\),又因为\(m\)是小于\(-1\)的整数,故\(l≤0\),
所以假设不成立,故\(m≥-1\).
②再证充分性:当\(a_1=md\),\(m≥-1\),\(m∈Z\),
任取等差数列\(\{a_n\}\)中不同的两项\(a_s\),\(a_t\),\((s≠t)\),
则\(a_s+a_t=2a_1+(s+t-2)d=a_1+(s+t+m-2)d\),
因为\(s+t+m-2≥0\)且\(s+t+m-2∈Z\),
所以\(a_1+(s+t+m-2)d=a_{s+t}+m-1\),
综上①②可得,数列中不同两项之和仍是这一数列中的一项的充要条件是存在整数\(m≥-1\),使\(a_1=md\).