设数列An是公差不为零且S1平方
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设数列{an}的公差为d,则各项分别为:a1,a1+d,a1+2d,…,a1+(n-1)d,且a1≠0,d≠0,假设去掉第一项,则有(a1+d)(a1+3d)=(a1+2d)2,解得d=0,不合题意;
设{A(n)}的通项公式为:A(n)=2+d(n-1){B(n)}的通项公式为:B(n)=2×q^(n-1)则{A(n)}的前n项和为:S(n)=[A(1)+A(n)]n/2=[4+d(n-1)]n/
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
设该等差数列首项a1,公差d则S1=a1S2=2a1+dS4=4a1+6d要成等比(2a1+d)^2=a1(4a1+6d)即4a1^2+4a1d+d^2=4a1^2+6a1d即d=2a1所以S1=a1
an=a1+(n-1)d=2+(n-1)da2=2+da4=2+3da8=2+7da2,a4,a8成等比数列,即a4/a2=a8/a4a4*a4=a2*a84+12d+9d^2=4+16d+7d^22
a3=b3a1+2d=b1*q^2=a1*q^2a1+2d=a1*q^2.1a7=b5a1+6d=b1*q^4=a1*q^4a1+6d=a1*q^4.21式×3-2式2a1=3a1*q^2-a1q^4
S1=a1S2=a1+a2=2a1+dS4=a1+a2+a3+a4=4a1+6d因为成等比数列,所以S2的平房=S1*S4(2a1+d)的平房=a1(4a1+6d)因为d不得0解得d=2a1所以S2=
S1=a1S2=a1+a2=2a1+dS4=a1+a2+a3+a4=4a1+6d因为成等比数列,所以S2的平房=S1*S4(2a1+d)的平房=a1(4a1+6d)因为d不得0解得d=2a1所以S2=
(1)∵数列{an}是公差不为零的等差数列,a1=2,且a2,a4,a8成等比数列,∴(2+3d)2=(2+d)(2+7d),解得d=2,∴an=2n.(2)∵an=2n,∴3an=32n=9n,此数
用求和公式,求解二元二次方程组.
由已知a3*a3=a2*a7S9=10,9*a5=90,有a5=10即(10-2d)^2=(10+2d)(10-3d)解得d=3所以A1=-2,d=3因此A2n=6n-5其前100项和为29800
设an=a+d*(n-1)1.a3+a10=a+2d+a+9d=2a+11d=152.a3*a7=a4*a4(a+2d)(a+6d)=(a+3d)^2a=-1.5d联立1与2,求得d=15/8a=-4
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
把首项和公差设出来解个二元一次方程组就行了设首项为a1公差为d则(1)[a1+(a1+d)+(a1+2d)]^2=9[a1+(a1+d)](2)a1+(a1+d)+(a1+2d)+(a1+3d)=4[
在等差数列中,公差d不为0,a11+40d=a51,即a11=a51-40d因为|a11|=|a51|,即a11=-a51,或者a11=a51(不符,舍去)所以a11+a51=2*a31=0,即a31
(1)因为a4,a5,a8成等比数列,所以a52=a4a8.设数列{an}的公差为d,则(3+3d)2=(3+2d)(3+6d)化简整理得d2+2d=0.∵d≠0,∴d=-2.于是an=a2+(n-2
a1a2a3成等比数列a2^2=a1a3=a3(a1+d)^2=a1+2da1^2+2a1d+d^2=a1+2d1+2d+d^2=1+2dd^2=0d=0公差不为零的等差数列错题
s1=a1s2=2a1+ds4=4a1+6d因为s1,s2,s4成等比数列所以(s2)²=s1×s4(2a1+d)²=a1(4a1+6d)4a1²+4a1d+d²
数列{an}是公差不为0的等差数列,设公差为d,S1,S2,S4成等比数列,则S22=S1•S4,∴( 2a1+d)2=a1•(4a1+6d),化简可得d=2a1∴a3a1=a1+2da1=
(1)设数列{an}的公差为d,由题意,得S22=S1•S4所以(2a1+d)2=a1(4a1+6d)因为d≠0所以d=2a1,故a2a1=3;(2)因为a5=9,d=2a1,a5=a1+8a1=9a