在等差数列an中 an不等于0

额..a1=da2=2dq=a2/a1=2d/d=2.

因为Sm/Sn=m^2/n^2,所以{[2a1+(m-1)d]*m}/{[2a1+(n-1)d]*n}=m^2/n^2,[2a1+(m-1)d]/[2a1+(n-1)d]=m/n,2a1n+(m-1)

原问题即：有两个数列,{An}{Bn},若Sm:Sn=m^2:n^2求Am:Bn(公差分别为d1,d2)Sm=A1d+0.5*m(m-1)d1=0.5d1m^2+m(A1-0.5d1)Sn=0.5d2

设公差为d,a1=-1,且其中a2,a4,a12三项成等比∴(-1+3d)^2=(-1+d)(-1+11d)解得d=0,3∵d≠0,∴公差为3a10=-1+3*9=26(2)a20=-1+3*19=5

设An=A1+(n-1)d则A2=A1+dA4=A1+3d因为A2是A1与A4等比中项故（A2）²=A1A4即（A1+d）²=A1（A1+3d）d²=A1d因为d不为0,

（a1+（cn-1）d）/(a1+((cn-1)-1)d)=(a1+2d)/a1=2利用合比得到cn-3=2（cn-1）-2这里的cn-1中n-1是下标递推式为cn=2cn-1+1两边同+1得到cn+

a5=a1+4d,a17=a1+16d因为a1,a5,a17成等比数列所以(a1+4d)^2=a1*(a1+16d)故(a1)^2+8a1*d+16d^2=(a1)^2+16a1*d即2d^2=a1*

a3=（a1+a5）/2b3=√b1b5=√a1a5∵a1≠a3,∴d≠0,a1≠a5a3-b3=（a1+a5-2√a1a5）/2=（√a1-√a5）²/2＞0∴a3＞b3再问：懂了

由题可知(b2）^2=b1xb3把a1=b1=1,a2=b2,a4=b3带入上式(a2）^2=a1xa4即（a1+d）^2=a1x（a1+3d）整理d=0舍掉或者d=a1=1则an=n所以b1=1b2

1.S5=5a1+10d=5(a1+2d)=70a1+2d=14a3=14a7^2=a2×a22(a3+4d)^2=(a3-d)(a3+19d)a3=14代入,整理,得d(d-4)=0d=0(已知d不

在等差数列{an}中,a1=10,公差为d,（1）由题意,S10=10a1+45d>0,得d>-20/9；S11=11a1+55d

设公比为q则由a3、a5,a6成等差数列得2a5=a3+a6即2*a1*q^4=a1*q^2+a1*q^5所以q^3-2q^2+1=0即(q-1)(q^2-q-1)=0解得q=(-1±√5)/2或q=

【解】（1）方程A(k)(X^2)+2A(k+1)X+A(k+2)=0,则其Δ=4[A(k+1)^2-A(k)*A(k+2)]=4[[A(k)+d]^2-A(k)*[A(k)+2d]]=4d^2>0;

a8+a14=2a1+20d=0a1=-10d0Sn=na1+n(n-1)d/2=-10nd+n^2d/2-nd/2=(d/2)*n^2-(21d/2)n,对称轴是n=21/2=10.5所以,当n=1

是等比数列吧?3a(n+1)-an=03a(n+1)=ana(n+1)/an=1/3,等比1/3a1=2an=2/3^(n-1)=6/3^n

（1）如果等比数列｛bn｝是递增的,则b(n+1)>bn对任意正整数n成立,若首项为b1,公比为q,则b1*q^n>b1*q^(n-1)对任意正整数n都成立,所以q>0,则b1>0时q>1,b1b1*

a(n)=1+(n-1)da(n+1)=1+ndSn=(1+an)n/2=(2+nd-d)n/2(1+Sn)/(n(1-a(n+1)))=-((4+nd-d)/n)/(2n(nd))=-2/(nd)-

a3^2+a8^2+2a3a8=9(a3+a8)^2=9因为等差数列an的各项都是负数所以a3+a8=-3所以S10=(a1+a10)*10/2=5(a1+a10)=5(a3+a8)=5*(-3)=-

am=a1+(m-1)d=xan=a1+(n-1)d=y两式相减得(m-n)d=x-yd=(x-y)/(m-n)代入a1+(m-1)d=x得a1=(my-nx+x-y)/(m-n)所以ai=a1+(i

证明：左边＝1/(a1a2)+1/(a2a3)+...+1/(an-1*an)＝1/d(1/a1-1/a2)+1/d(1/a2-1/a3)+...+1/d(1/an-1-1/an)=1/d[(1/a2