已知等差数列an的公差d≠0,它的前几项和为Sn,若S5=70

a1,a3,a4成等比数列所以(a1+2d)^2=a1*(a1+3d)a1^2+4d*a1+4d^2=a1^2+3d*a1所以d*a1=-4d^2因为d≠0所以：a1=-4da5=a1+4d=0

∵an为等差数列a1,a3,a9成等比数列∴a1(a1+8d)=(a1+2d)^2a1^2+8d*a1=a1^2+4d*a1+4d^2d≠0∴d=a1a1+a3+a9/a2+a4+a10=(a1+a1

显然有：an=a1+(n-1)d,bn=b1*q^(n-1),又a3=b3,a7=b5,所以：a1+2d=a1*q^2,①a1+6d=a1*q^4,②由上面2个式子,得到：3①-②：2a1=a1*(3

a1^2=a11^2,∴a1=-a11a1=-(a1+10d)2a1=-10da1=-5dan=a1+(n-1)d=-5d+(n-1)d=(n-6)d∵d0,a6=0,a7

a1,a3,a9成等比数列a3^2=a1*a9(a1+2d)^2=a1*(a1+8d)解得a1=d(a1+a3+a9)/(a2+a4+a10)=(3a1+10d)/(3a1+13d)=13d/16d=

a2+a4=2*a3=8a3=4,a4=3因此a1=6,d=-1通项为an=6-(n-1)=7-n

a1,a5,a17成等比数列a5^2=a1a17(a1+4d)^2=a1(a1+16d)a1^2+8a1d+16d^2=a1^2+16a1d16d^2=8a1da1=2d所以/=/=/=/=26/29

首项为a1,公差为d,a1,a1+4d,a1+16d成等比数列,则(a1+4d)^2=a1*(a1+16d)得d=0,或a1=2d,当d=0时,/=1当d=0时,/=(3a1+20d)/(3a1+23

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

是等差数列设首相是a1那么an=a1+（n-1）dakn=a1+(kn-1)dak（n+1）=a1+（k（n+1）-1）d-（a1+(kn-1)d）=kd所以{akn}是等差数列2）已知等比数列{bn

等差数列{an}公差为d(d≠0),前n项和Sn=na1+n(n-1)d/2∴xn=Sn/n=a1+(n-1)d/2∴{xn}为等差数列,首项为a1公差为d/2∴｛Xn｝的前n项和Tn=n[2a1+(

ak=48+2kbk=10+(k-1)dSk=(48+2k)[10+(k-1)d]令SK≤21即(48+2k)[10+(k-1)d]≤21求出k来.再问：最大圆面积为Sk

a1,a5,a17是等比数列(a1+4d)^2=a1*(a1+16d)a1^2+8a1d+16d^2=a1^2+16a1d8a1d=16d^2d不等于0a1=2dq=a5/a1=(a1+4d)/a1=

因为a(k1),a(k2),…,a(kn)恰为等比数列,又k1=1,k2=5,k3=17所以a5的平方=a1乘以a17又因为数列｛an｝为等差数列且公差d≠0所以a5=a1+4da17=a1+16d所

因为{An}是等差数列,所以A2+A8=A4+A6=10,A4*A6=24,所以可将A4、A6看作方程x^2-24x+10=0的两个根,因为d

由题可得A1*A9等于A3方把分子分母都写为A3和公差d的表达式有上式可得A3和d的关系带入就可的到比值

a2,a5,a14是等比数列所以（a5)^2=a2*a14即（a+4d)^2=(a+d)*(a+13d)化简得d=2a所以公比q=a5/a2=(a+4*2a)/(a+2a)=3(2)a122=a+12

a3=3b3a1+2d=3a1d²(3d²-1)a1=2da1=2d/(3d²-1)a5=5b5a1+4d=5a1d⁴(5d⁴-1)a1=4da1

再问：太给力了，你的回答完美解决了我的问题！

先求An的通项就行了A1+A4=14A2A3=45d