设公差不为0的等差数列,a2平方 a3平方=a4平方 a5平方,S7=7

由于为等比数列,只要连续3项就可确定数列的首项和公比!故只需要讨论4项删去某一项后剩3项即可!故只要讨论a1,a2,a3,a4即可!(1)删掉首项:a2,a3,a4a3^2＝(a3-d)(a3+d)d

请参考下图：

(1)令通项公式:an=a1+(n-1)da2=a1+da4=a1+3dS10=5(2a1+9d)=110由题意:a2^2=a1*a4即(a1+d)^2=a1*(a1+3d)由题意:a1=d=2所以通

先给出答案：a1/a2＝1/3序号第n项前n项和Sn第1项：aa第2项：a+d2a+d第3项：a+2d3a+3d第4项：a+3d4a+6dS1：S2＝S2：S4或者(S2)^2=＝S1*S4(2a+d

a2=a1+da4=a1+3da6=a1+5da2,a4－2,a6成等【比】数列(a1+3d-2)^2=(a1+d)(a1+5d)(3d-1)^2=(1+d)(1+5d)9d^2-6d+1=5d^2+

设公差da2=a1+da4=a1+3da10=a1+9dS10=(a1+a10)*10/2=5(2a1+9d)=1102a1+9d=22a1.a2.a4为等比数列a2*a2=a1*a4(a1+d)^2

a3=a2+d=2+d;a5=a2+3d=2+3d;所以2（2+3d）=(2+d)²;4+6d=4+4d+d²;d²-2d=0;d=0或d=2；因为公差不等于0；所以d=

1.若n=4时,则原数列为a1,a2,a3,a4.⑴若删去a1,则a3∧2=a2×a4,→d=0,矛盾⑵若删去a2,→a5=0矛盾⑶若删去a3→a1=d→a1/d=1⑷若删去a4→d=0矛盾综上所述,

答案已给出

易知（a2+d)^2=a2*(a2+4d)得：d=2a2所以(a1+a3+a5)/(a2+a4+a6)=（a2-d+a2+d+a2+3d)/(a2+a2+2d+a2+4d)=（a2+a2+a2+6a2

设{an}是一个公差为d(d≠0)的等差数列,它的前10项s10=110且a1,a2,a4成等比数列.a1*a4=a2^2a1*(a1+3d)=(a1+d)^2a1=d或d=0(舍去）an=d*nsn

你的答案似乎不对,因为我做过这道题三遍.1.a1,a3,a5成等比则：a3^2=a1*a5又a1,a3,a5是等差数列{an}中的项则:a3=a1+2da5=a1+4d则有:(a1+2d)^2=a1(

1.a1a2=a1+da4=a1+3da2^2=a1*a4(a1+d)^2=a1*(a1+3d)2a1*d+d^2=3a1*dd=a1S10=110=10a1+10*9*d/2a1=d=2{an}=2

首项为a1,公差为dS10=10a1+45d=110.(1)a1,a2,a4成等比数列.(a2)^2=a1*a4(a1+d)^2=a1(a1+3d).(2)通过(1)(2)得a1=d=2an=a1+(

（I）由a1，a2，a4成等比数列可得：(a1+2)2＝a1(6+a1)∴4=2a1即a1=2∴an=2+2（n-1）=2n（II）∵bn=n•2an，=n•22n=n•4n∴Sn＝1•4+2•42+

a2=a1+da4=a1+3da6=a1+5da2,a4－2,a6成等【比】数列(a1+3d-2)^2=(a1+d)(a1+5d)(3d-1)^2=(1+d)(1+5d)9d^2-6d+1=5d^2+

设该等差数列是首项为a1,公差为dS3=3a1+3(3-1)*d/2=3a1+3dS2=2a1+2(2-1)*d/2=2a1+dS4=4a1+4(4-1)*d/2=4a1+6d又:S3²=9

（1）设等差数列{an}的公差为d，由a22＝a1a4，…（1分）得(a1+d)2＝a1(a1+3d)…（2分）∵d≠0，∴d=a，∴an=na1，Sn＝an(n+1)2．（2）∵1Sn＝2a(1n−

a2=a1+da3=a1+2da6=a1+5d由等比数列性质（a1+2d）^2=(a1+d)(a1+5d)a1=-1/2dq=a3/a2=3

A1=1,A2=1+d,A8=1+7d;B1=1,B2=1*q,B3=1*q^2=>1+d=q;1+7d=q^2=>d=5,q=6,A2=B2=6,A8=B3=36S(Bn)=A1(1-q^n)/(1