4．设Sn是等差数列{an}的前n项和，若S3S6＝13，则S6S12等于()A.310B.13C.

设Sn是等差数列{an}的前n项和,若S7＝35,则a4等于A.8B.7C.6D.5答案DSn是等差数列{an}的前n项和,若S7＝7a4＝35,∴a4＝5,选D．查看原帖>>

Sn是等差数列｛an｝的前n项和,已知a2=3,a6=11则S7等于几选项A13B35C49D63a2+a6=a1+a7S7=(a1+a7)*7/2=(a2+a6)*7/2=49这里有个关系是Am+Am=Ap+Aq条件为m+n=p+qa6=

设Sn是等差数列{an}的前n项和,若S3/S6=1/4,则S6/S12等于?设等差数列的首项为a1,公差为d.则有（a1+2d）/(a1+5d)=1/4,解得d=-a1那么S6/S12=（a1+5d）/(a1+11d)=(-4a1)/(-

设Sn是等差数列{an}的前n项和,若a5/a3=5/9,则S9/S5=：A1B-1C2D1/2请解释为什么S9/S5=(9a5)/(5a3)=1选A

设Sn是等差数列an的前n项和,若S7=a5+a9,则a4\a7?A1\3B2\7C1\2D1s7=(a1+a7)*7/2=7a4=a5+a9=2a7,所以a4\a7=2/7,B

设等差数列an的前n项和为Sn,若Sn＝3,S3n＝15,则S5n等于?n是项数,d是公差Sn=3,S2n=2Sn+dn^2,S3n=3Sn+3dn^2,S4n=4Sn+6dn^2,S5n=5Sn+10dn^2.因为S3n=15=3Sn+3

若数列An的前n项和为Sn=an^2+bn+c,(a,b,c属于正整数）则An为等差数列的充要条件是c=0.为什么证明：充分性：当c=0时：an=Sn-Sn-1=(2n-1)a+b(n≥2)a1=S1=a+b+c所以当c=0时a1=a+b所

设Sn是等差数列An的前n项和,若S3/S6=1/3,则S6/S12等于等差数列S3,S6-S3,S9-S6,S12-S9也成等差数列S3/S6=1/3,S6=3S3,S6-S3=2S3S9-S6=3S3,S9=6S3S12-S9=4S3,

设数列{an}是等差数列且a4=-4,a9=4,Sn是数列{an}的前n项和,则AS5＜S6BS5=S6CS7=S5DS7=S6选c（1）a4+a9=a6+a7=-4+4=0,所以s7=s5+a6+a7=s5+0=s5所以选（c）（2）由a

设Sn为等差数列{An}的前n项和,求证：{Sn/n}是等差数列Sn=(a1+an)*n/2所以Sn/n=(a1+an)/2=[a1+a1+(n-1)d]/2=a1+(d/2)*(n-1)这是以a1为首项,d/2为公差的等差数列所以{Sn/

设Sn是等差数列(an)的前n项和,已知a2等于3,a6等于11,则S7等于(1)S7=(a1+a7)*7/2=(a2+a6)/7*2=(11+3)*7/2=49(2)d=(a6-a2)/(6-2)=(11-3)/4=2a1=a2-d=1a

设数列{an}是等差数列,且a2等于-6,a8等于6,Sn是数列｛an｝的前n项和则A.S4S5D.S6=S5哪个对?可是按照你这样算C不也对了么？2楼貌似列错式子了..a2等于-6,a8等于6a1+d=-6a1+7d=6d=2,a1=-8

设sn是等差数列an的前n项和,若a20120,则使sn>0的最小正整数值n为A2013B2014C4023D4024a20120zuix则a1+a4023=2a20120所以n最小是4024

等差数列前n项求和设Sn是等差数列{an}的前n项和,若a5\a3=5\9,则S9\S5=()A.1B.-1C.2D.1\2求详解啊～谢谢S9/S5=((a1+a9)*9)/((a1+a5)*5)=(2*a5*9)/(2*a3*5)=9*a

等差数列〔an〕的前n项和为Sn,已知Sm等于a,Sn减Sn-m等于b,n,m是正的等差数列〔an〕的前n项和为Sn，已知Sm等于aSn减Sn-m等于bnm是正的自然数。n大于m求Sn由：Sm=a,及b可求a1；由：Sn=Sn-m+Sm+(

已知等差数列an的前n项和为sn,若a3+a7+a11=6,则S13等于()A.24B.25C.26D.27a3+a7+a11=3a7=6/a3+a11=2a7,等差中项性质a7=2S13=(a1+a13)·13/2=2a7·13/2=13

等差数列{an}的前n项和为Sn,且S3=6,a1=4,则公差d等于A.1B.5/3C.-2D.3选c啊首项为4第二项为2第三项为0

设{an}与{bn}是两个等差数列,它们的的前n项和分别为Sn和Tn,若Sn/Tn＝3n+1/4n-3,那么an/bn等于多少不妨利用原理：a(m)+a(n)＝2a[(m+n)/2]（以上m,n都表示项数）可以知道：S(2n-1)＝a(1)

设等比数列｛an｝的公比为q,前n项和为Sn,若Sn+1,Sn,Sn+2成等差数列,则q等于多少?若an=1,求sn前n项和tn(1)∵｛An｝为等比数列,则有An+1=An·q,又∵Sn+1,Sn,Sn+2成等差数列,∴Sn+1+Sn+2

设两个等差数列{an},{bn}的前n项和分别为Sn,Tn.若Sn/Tn=7n+1/4n+27,则a7/b7=取N=13Sn=(a1+a13)X13/2Tn=(b1+b13)x13/2a1+a13=2Xa7同理b7则比值为92/79