a3=3,a7=1有数列an加1分之1s是等差数列
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没得选-1/5a3+a5+a7=6,=>a5=2,=>a7+a8=2a5+5d=3,=>d=-1/5
因为公比q=-1/3a1+a3+a5+a7是以a1为首项,公比是1/9的等比数列a2+a4+a6+a8是以a1q为首项,公比是1/9的等比数列所以(a1+a3+a5+a7)/(a2+a4+a6+a8)
a2+a4+a6+a8=a1q+a3q+a5q+a7q=q(a1+a3+a5+a7)所以a1+a3+a5+a7/a2+a4+a6+a8=1/q=3选D
由a1+a2=a1(1+q)=3①,a2+a3=a1q(1+q)=6②,②÷①得:q=2,把q=2代入①得到a1=1,则a7=26=64.故答案为:64
==等比数列的性质没记住啊a1a7=a4²a2a6=a4²a3a5=a4²所以a1*a2*a3……a7=a4^7=3^7=2187
∵a3+a4=(a1+2d)+(a2+2d)=a1+a2+4d∴(a3+a4)-(a1+a2)=(a1+a2+4d)-(a1+a2)=6-3=3=4d∴d=3/4∴a7+a8=(a3+4d)+(a4+
因为a1*a10=3所以a2*a9=3a3*a8=3a4*a7=3a5*a6=3所以a2*a3*a4*a5*a6*a7*a8*a9=3*3*3*3=81
错了an=a1q^(n-1)这才是等比数列的通项公式原式=a1(1+q²+q^4+q^6)/a1(q+q^3+q^5+q^7)=1/q=-3
a1=3/8a2=-1/8a3=1/24a4=-1/27a5=1/216a6=-1/648a7=1/1944a8=-1/5832a9=1/17496a10=-1/52488等比数列的比值为-1/3
由等差数列的性质得:3a5+a7=2a5+(a5+a7)=2a5+(2a6)=2(a5+a6)=2(a3+a8)=20,故答案为:20.
Ca3=—9=a1q^2a7=—1=a1q^6用2式除以1式得q^4=1/9所以q^2=1/3由1式子,a1=-27所以a5=a1×q^4=-27×1/9=-3
解析a1q^2=1/2a1q^6=2所以q^4=4q=±√2a1=1/4a5=a1q^4=1/4x4=1
(1)a1(1+q^6)=65a1^2*q^6=64由a(n+1)
4因为a3=a1+2d(d为等差)a4=a1+3da5=a1+4da7=a1+6da10=a1+9da13=a1+12d全部换过来的话3(a3+a5)+2(a7+a10+a13)=24=》a1+6d=
∵a7-2a4=-1,且a3=0,∴a3+4d-2(a3+d)=-1,解得d=−12.故选B.
sin²a3=(1/2)(1-cos2a3)sin²a7=(1/2)(1-cos2a7)sin(a3+a7)=sin2a5于是cos2a7-cos2a3=-2sin2a52sin(
dn=n次根下(a1*a2*a3*a4.*an)还有,怎么出来cn了
a5(a3+2a5+a7)=a5a3+2a5a5+a5a7=a4a4+2a4a6+a6a6=(a4+a6)(a4+a6)=9
设an=a1+(n-1)d-1=(sin^2a3-sin^2a7)/sin(a3a7)=(sina3+sina7)(sina3-sina7)/sin(a3a7)=4sina5cos(2d)sin(-2