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The sequence \(a_1\), \(a_2\), \(a_3\), ... , \(a_n\), ... is such that \(a_n=\frac{a_{n-1}+a_{n-2}}{2}\) for all \(n\geq{3}\). If \(a_3 = 4\) and \(a_5 = 20\), what is the value of \(a_6\) ?(A) 12(B) 16(C) 20(D) 24(E) 28Diagnostic TestQuestion: 3Page: 20Difficulty: 600
Spoiler: OA
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SOLUTIONThe sequence \(a_1\), \(a_2\), \(a_3\), ... , \(a_n\), ... is such that \(a_n=\frac{a_{n-1}+a_{n-2}}{2}\) for all \(n\geq{3}\). If \(a_3 = 4\) and \(a_5 = 20\), what is the value of \(a_6\) ?(A) 12(B) 16(C) 20(D) 24(E) 28Since given that \(a_n=\frac{a_{n-1}+a_{n-2}}{2}\), then:\(a_5=\frac{a_{4}+a_{3}}{2}\) --& \(20=\frac{a_{4}+4}{2}\) --& \(a_4=36\);\(a_6=\frac{a_{5}+a_{4}}{2}\) --& \(a_6=\frac{20+36}{2}\) --& \(a_5=28\).Answer: E.
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Hi,This one has be solved step by step, and there is chance of making mistakes.Difficulty level: 600\(a_n=\frac{a_{n-1}+a_{n-2}}2\)or a_n is the average of last two terms,thus,\(\frac{a_3+a_4}2=a_5\)\(\frac{a_4+a_5}2=a_6\)Subtrac\(\frac{a_5-a_3}2=a_6-a_5\)\(a_6=\frac{3a_5-a_3}2=\frac{3*20-4}2=28\)Thus, Answer is (E).Regards,
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a4=2*a6-a5;=&a4=40-4=36;
therefore,a6=(36+20)/2=28;IF U GUYS LIKE THE POST PLZ GIVE A KUDOS!!!!
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a5 = (a3+a4) / 2=& a4 = 2(a5) - a3
= 2 x 20 - 4 = 36a6 = (a4+a5) / 2
= (36 + 20) / 2
= 28Option E is correct.Difficulty level is 600.
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Its Efrom the given formula we can deduce A4 and then take average of A4 and A5 and there is your answer
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its E.a5=(a4+a3)/2substituting you get a4=36now a6=(a5+a4)/2substitute to get a6 as 28
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I'm not sure how efficient my method was:a3 = 4 = (a2 + a1) / 2:. 8 = a2 + a1a5 = 20 = (a4 + a3) / 2:. 40 = a4 + a3 = a4 + 4:. a4 = 36a6 = ?a6 = (a5 + a4) / 2a6 = (20 + 36) / 2 = 28(E) 28
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I consider it to be higher then 600 level due to function involved although very each one if one is able to decipher. Calculate the value of a4 as a5 and a3 is given. Thus a5=a4+a3/220=(a4+4)/2 or
40-4=a4. a6= 20+36/2= 28Answer: E
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\(a_3 = 4\)\(a_4 = ?\)\(a_5 = 20\)\(a_6 = ?\)\(a_n=\frac{a_{n-1} + a_{n-2}}{2}\)\(20(2)=\frac{4 + a_4}{2}\)\(40=4 + a_4\)\(36 = a_4\)\(a_6 = \frac{20 + 36}{2}=28\)Answer: (E) 28
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Bunuel wrote:The sequence \(a_1\), \(a_2\), \(a_3\), ... , \(a_n\), ... is such that \(a_n=\frac{a_{n-1}+a_{n-2}}{2}\) for all \(n\geq{3}\). If \(a_3 = 4\) and \(a_5 = 20\), what is the value of \(a_6\) ?(A) 12(B) 16(C) 20(D) 24(E) 28Diagnostic TestQuestion: 3Page: 20Difficulty: 600Nice but easy problem. I think the idea is based on Fibonacci sequence.
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Bunuel wrote:The sequence \(a_1\), \(a_2\), \(a_3\), ... , \(a_n\), ... is such that \(a_n=\frac{a_{n-1}+a_{n-2}}{2}\) for all \(n\geq{3}\). If \(a_3 = 4\) and \(a_5 = 20\), what is the value of \(a_6\) ?(A) 12(B) 16(C) 20(D) 24(E) 28Please note:
this question in the 2017 version of the
(page 20, #3, Quant Diagnostic Test) contains a typo.
It should say \(a_n=\frac{a_{n-1}+a_{n-2}}{2}\) for all \(n\geq{3}\), but instead it says \(a_n=\frac{a_{n+1}+a_{n-2}}{2}\) for all \(n\geq{3}\).
The answer explanation on page 46, however, lists the correct formula.Yes, even the GMAC makes mistakes!
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Here A5=A4+A3/2 =& A4=36 A6=36+20/2 =& 18+10 =& 28SMASH that E
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a5 = 20a5 = \(\frac{a4 + a3}{2}\)20 =\(\frac{a4 + 4}{2}\)40 - 4 = a4a4 = 36a6 = \(\frac{20 + 36}{2}\)28E
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Option E\(a_n=\frac{a_{n-1}+a_{n-2}}{2}\)a\(_5 = 20, a_3 = 4\)\(a_5=\frac{a_{4}+a_{3}}{2}\)\(a_4 = 36\)\(a_6=\frac{a_{5}+a_{4}}{2}\) = 28
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by Problem Solving (PS)& 等差数列与等比数列的综合知识点 & “已知集合Sn={（x1，x2，…，xn）...”习题详情
0位同学学习过此题，做题成功率0%
已知集合Sn={（x1，x2，…，xn）|x1，x2，…，xn是正整数1，2，3，…，n的一个排列}（n≥2），函数对于（a1，a2，…an）∈Sn，定义：bi=g（ai-a1）+g（ai-a2）+…+g（ai-ai-1），i∈{2，3，…，n}，b1=0，称bi为ai的满意指数．排列b1，b2，…，bn为排列a1，a2，…，an的生成列；排列a1，a2，…，an为排列b1，b2，…，bn的母列．（Ⅰ）当n=6时，写出排列3，5，1，4，6，2的生成列及排列0，-1，2，-3，4，3的母列；（Ⅱ）证明：若a1，a2，…，an和a′1，a′2，…，a′n为Sn中两个不同排列，则它们的生成列也不同；（Ⅲ）对于Sn中的排列a1，a2，…，an，定义变换τ：将排列a1，a2，…，an从左至右第一个满意指数为负数的项调至首项，其它各项顺序不变，得到一个新的排列．证明：一定可以经过有限次变换τ将排列a1，a2，…，an变换为各项满意指数均为非负数的排列．&
本题难度：一般
题型：解答题&|&来源：2013-北京市石景山区高考数学二模试卷（理科）
分析与解答
习题“已知集合Sn={（x1，x2，…，xn）|x1，x2，…，xn是正整数1，2，3，…，n的一个排列}（n≥2），函数对于（a1，a2，…an）∈Sn，定义：bi=g（ai-a1）+g（ai-a2）+…+g（ai...”的分析与解答如下所示：
当n=6时，排列3，5，1，4，6，2的生成列为0，1，-2，1，4，-3；排列0，-1，2，-3，4，3的母列为3，2，4，1，6，5．（Ⅱ）证明：设a1，a2，…，an的生成列是b1，b2，…，bn；a′1，a′2，…，a′n的生成列是与b′1，b′2，…，b′n，从右往左数，设排列a1，a2，…，an与a′1，a′2，…，a′n第一个不同的项为ak与a′k，即：an=a′n，an-1=a′n-1，…，ak+1=a′k+1，ak≠a′k．显然&bn=b′n，bn-1=b′n-1，…，bk+1=b′k+1，下面证明：bk≠b′k．由满意指数的定义知，ai的满意指数为排列a1，a2，…，an中前i-1项中比ai小的项的个数减去比ai大的项的个数．由于排列a1，a2，…，an的前k项各不相同，设这k项中有l项比ak小，则有k-l-1项比ak大，从而bk=l-（k-l-1）=2l-k+1．同理，设排列a′1，a′2，…，a′n中有l′项比a′k小，则有k-l′-1项比a′k大，从而b′k=2l′-k+1．因为&a1，a2，…，ak与a′1，a′2，…，a′k是k个不同数的两个不同排列，且ak≠a′k，所以&l≠l′，从而&bk≠b′k．所以排列a1，a2，…，an和a′1，a′2，…，a′n的生成列也不同．（Ⅲ）证明：设排列a1，a2，…，an的生成列为b1，b2，…，bn，且ak为a1，a2，…，an中从左至右第一个满意指数为负数的项，所以&b1≥0，b2≥0，…，bk-1≥0，bk≤-1．进行一次变换τ后，排列a1，a2，…，an变换为ak，a1，a2，…ak-1，ak+1，…，an，设该排列的生成列为b′1，b′2，…，b′n．所以&（b′1，b′2，…，b′n）-（b1+b2+…+bn）=[g（a1-ak）+g（a2-ak）+…+g（ak-1-ak）]-[g（ak-a1）+g（ak-a2）+…+g（ak-ak-1）]=-2[g（ak-a1）+g（ak-a2）+…+g（ak-ak-1）]=-2bk≥2．因此，经过一次变换τ后，整个排列的各项满意指数之和将至少增加2．因为ai的满意指数bi≤i-1，其中i=1，2，3，…，n，所以，整个排列的各项满意指数之和不超过1+2+3+…+（n-1）=，即整个排列的各项满意指数之和为有限数，所以经过有限次变换τ后，一定会使各项的满意指数均为非负数．
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已知集合Sn={（x1，x2，…，xn）|x1，x2，…，xn是正整数1，2，3，…，n的一个排列}（n≥2），函数对于（a1，a2，…an）∈Sn，定义：bi=g（ai-a1）+g（ai-a2）+…...
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经过分析，习题“已知集合Sn={（x1，x2，…，xn）|x1，x2，…，xn是正整数1，2，3，…，n的一个排列}（n≥2），函数对于（a1，a2，…an）∈Sn，定义：bi=g（ai-a1）+g（ai-a2）+…+g（ai...”主要考察你对“等差数列与等比数列的综合”
等考点的理解。
因为篇幅有限，只列出部分考点，详细请访问。
等差数列与等比数列的综合
等差数列与等比数列的综合．
与“已知集合Sn={（x1，x2，…，xn）|x1，x2，…，xn是正整数1，2，3，…，n的一个排列}（n≥2），函数对于（a1，a2，…an）∈Sn，定义：bi=g（ai-a1）+g（ai-a2）+…+g（ai...”相似的题目：
已知数列{an}中，Sn是它的前n项和，并且Sn+1=4an+2，a1=1．（1）设bn=an+1-2an，求证{bn}是等比数列（2）设，求证{Cn}是等差数列（3）求数列{an}的通项公式及前n项和公式&&&&
已知等差数列{an}的公差为-1，且a2+a7+a12=-6，（1）求数列{an}的通项公式an与前n项和Sn；（2）将数列{an}的前4项抽去其中一项后，剩下三项按原来顺序恰为等比数列{bn}的前3项，记{bn}的前n项和为Tn，若存在m∈N*，使对任意n∈N*总有Sn＜Tm+λ恒成立，求实数λ的取值范围．&&&&
设{an}是由正数组成的等差数列，{bn}是由正数组成的等比数列，且a1=b1，a2003=b2003，则必有&&&&a1002＞b1002a1002=b1002a1002≥b1002a1002≤b1002
“已知集合Sn={（x1，x2，…，xn）...”的最新评论
该知识点好题
1已知{an}是公差为d的等差数列，{bn}是公比为q的等比数列．（1）若an=3n+1，是否存在m、k∈N*，有am+am+1=ak？说明理由；（2）找出所有数列{an}和{bn}，使对一切n∈N*，an+1an=bn，并说明理由；（3）若a1=5，d=4，b1=q=3，试确定所有的p，使数列{an}中存在某个连续p项的和是数列{bn}中的一项，请证明．
2已知等比数列{an}的前n项和为Sn=ao2n+b，且a1=3．（1）求a、b的值及数列{an}的通项公式；（2）设bn=nan，求数列{bn}的前n项和Tn．
3已知等差数列{an}的公差为-1，且a2+a7+a12=-6，（1）求数列{an}的通项公式an与前n项和Sn；（2）将数列{an}的前4项抽去其中一项后，剩下三项按原来顺序恰为等比数列{bn}的前3项，记{bn}的前n项和为Tn，若存在m∈N*，使对任意n∈N*总有Sn＜Tm+λ恒成立，求实数λ的取值范围．
该知识点易错题
1已知{an}是公差为d的等差数列，{bn}是公比为q的等比数列．（1）若an=3n+1，是否存在m、k∈N*，有am+am+1=ak？说明理由；（2）找出所有数列{an}和{bn}，使对一切n∈N*，an+1an=bn，并说明理由；（3）若a1=5，d=4，b1=q=3，试确定所有的p，使数列{an}中存在某个连续p项的和是数列{bn}中的一项，请证明．
2已知等比数列{an}的前n项和为Sn=ao2n+b，且a1=3．（1）求a、b的值及数列{an}的通项公式；（2）设bn=nan，求数列{bn}的前n项和Tn．
3已知等差数列{an}的公差为-1，且a2+a7+a12=-6，（1）求数列{an}的通项公式an与前n项和Sn；（2）将数列{an}的前4项抽去其中一项后，剩下三项按原来顺序恰为等比数列{bn}的前3项，记{bn}的前n项和为Tn，若存在m∈N*，使对任意n∈N*总有Sn＜Tm+λ恒成立，求实数λ的取值范围．
欢迎来到乐乐题库，查看习题“已知集合Sn={（x1，x2，…，xn）|x1，x2，…，xn是正整数1，2，3，…，n的一个排列}（n≥2），函数对于（a1，a2，…an）∈Sn，定义：bi=g（ai-a1）+g（ai-a2）+…+g（ai-ai-1），i∈{2，3，…，n}，b1=0，称bi为ai的满意指数．排列b1，b2，…，bn为排列a1，a2，…，an的生成列；排列a1，a2，…，an为排列b1，b2，…，bn的母列．（Ⅰ）当n=6时，写出排列3，5，1，4，6，2的生成列及排列0，-1，2，-3，4，3的母列；（Ⅱ）证明：若a1，a2，…，an和a′1，a′2，…，a′n为Sn中两个不同排列，则它们的生成列也不同；（Ⅲ）对于Sn中的排列a1，a2，…，an，定义变换τ：将排列a1，a2，…，an从左至右第一个满意指数为负数的项调至首项，其它各项顺序不变，得到一个新的排列．证明：一定可以经过有限次变换τ将排列a1，a2，…，an变换为各项满意指数均为非负数的排列．”的答案、考点梳理，并查找与习题“已知集合Sn={（x1，x2，…，xn）|x1，x2，…，xn是正整数1，2，3，…，n的一个排列}（n≥2），函数对于（a1，a2，…an）∈Sn，定义：bi=g（ai-a1）+g（ai-a2）+…+g（ai-ai-1），i∈{2，3，…，n}，b1=0，称bi为ai的满意指数．排列b1，b2，…，bn为排列a1，a2，…，an的生成列；排列a1，a2，…，an为排列b1，b2，…，bn的母列．（Ⅰ）当n=6时，写出排列3，5，1，4，6，2的生成列及排列0，-1，2，-3，4，3的母列；（Ⅱ）证明：若a1，a2，…，an和a′1，a′2，…，a′n为Sn中两个不同排列，则它们的生成列也不同；（Ⅲ）对于Sn中的排列a1，a2，…，an，定义变换τ：将排列a1，a2，…，an从左至右第一个满意指数为负数的项调至首项，其它各项顺序不变，得到一个新的排列．证明：一定可以经过有限次变换τ将排列a1，a2，…，an变换为各项满意指数均为非负数的排列．”相似的习题。百度知道 - 信息提示
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已知等差数列{an}的公差d∈（0，1），且3+a7)sin(a3-a7)sina5cosa5=-2，当n=10时，数列{an}的前n项和Sn取得最小值，则首项a1的取值范围为（　　）A．[-，-]B．[-，-）C．（-，-]D．（-，-）
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由3+a7)sin(a3-a7)sina5cosa5=-2，得sin（a7+a3）sin（a7-a3）=sin（2a5），∵数列{an}是等差数列，∴sin（a7+a3）=sin（2a5）于是sin（a7-a3）=1，即sin4d=1，0＜d＜1于是d=．∵当n=10时，数列{an}的前n项和Sn取得最小值，∴10＝a1+9d＜0a11＝a1+10d＞0，解得-10d＜a1＜-9d．∴1＜-9π8．即a1的取值范围为．故选：D．
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把已知的等式变形，得到sin4d=1，结合d的范围求出d的值，再由当n=10时，数列{an}的前n项和Sn取得最小值得到10＝a1+9d＜0a11＝a1+10d＞0，求出a1的范围后代入d的值得答案．
本题考点：
等差数列的前n项和；三角函数的化简求值．
考点点评：
本题考查了三角函数的化简求值，考查了等差数列的通项公式，考查了不等式组的解法，是中档题．
扫描下载二维码在等比数列{an}中,a5=1,a10=3,则a4xa5x……xa11= 要过程！！_百度知道
在等比数列{an}中,a5=1,a10=3,则a4xa5x……xa11= 要过程！！
在等比数列中，有性质：an*an=ap*aq
(m+n=p+q),所以a4a5a6a7a8a9a10a11=(a4a11)(a5a10)(a6a9)(a7a8)=(a5a10)(a5a10)(a5a10)9a5a10)=(a5a10)^4=(1*3)^4=3^4=81
采纳率：40%
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