若都是等差数列,且公差相等,若a1,a2,a5恰为等比数列

设公差为d∵等差数列{an}的首项为25,且s9=s17∴9a1+1/2×9×8×d=17a1+1/2×17×16×d∴d=-2a1=25,d=-2∴an=a1+（n-1）d=27-2n再答：Sn=-

设等差数列的公差为d，首项为a1，所以a3=a1+2d，a7=a1+6d．因为a1、a3、a7成等比数列，所以（a1+2d）2=a1（a1+6d），解得：a1=2d．所以a1a4=2d5d=25．故选

就是Sn是等差数列若q≠1Sn=a1(1-q^n)/(1-q)则2Sn=S(n-1)+S(n+1)所以2a1(1-q^n)/(1-q)=a1[1-q^(n-1)]/(1-q)+a1[1-q^(n+1)

设等差数列的公差为d,a1,a3,a13成等比数列,则(a3)²=a1•a13(1+2d)²=1+12d,4d²=8d.因为公差不为0,所以d=2.从而an=

设公差为d由题意,S1=a1S2=2a1+dS4=4a1+6da1*(4a1+6d)=(2a1+d)^2d^2-2a1*d=0d(d-2a1)=0d不等于0所以d=2a1(1)公比q=S2/S1=(2

设等差数列的公差为d，首项为a1，所以a3=a1+2d，a7=a1+6d．因为a1、a3、a7成等比数列，所以（a1+2d）2=a1（a1+6d），解得：a1=2d．所以a2a1＝3d2d=32．故选

a3/a1=a4/a3即为：（a1+2d）/a1=（a1+3d）/（a1+2d）因为d=2,即为a1²+,8a1+16=a1²+6a1即得a1=-8故a2=-8+2=-6

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

a1+a3+a5=-12,即3a3=-12,a3=-4a3a4a5=18,得a4a5=-9/2,即(a3+d)(a3+2d)=-9/2(-4+d)(-4+2d)=-9/216-12d+2d^2=-9/

{an},{√sn}都是等差数列,∴2√S2=√S1+√S3,即2√（2a1+d)=√a1+√（3a1+3d),平方得4(2a1+d)=a1+3a1+3d+2√[a1(3a1+3d)],4a1+d=2

因为a2、a3、a6成等比数列，所以a32=a2•a6⇒（a1+2d）2=（a1+d）（a1+5d）⇒2a1d+d2=0．∵d≠0，∴d=-2a1．∴q=a3a2＝a1+2da1+d=3．故选C．

Tn=b1*b2*b3*……*bn=b1*(b1*q)*(b1*q^2)*……*[b1*q^(n-1)]=(b1)^n*q^[1+2+……+(n-1)]=(b1)^n*q^[n(n-1)/2]={b1

0,2,4,6a1=0a20=a1+19d=0+19x2=38Sn=20x(a1+a20)/2=10x(0+38)=380

∵|a3|=|a9|且d0a9

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

设：a1=b1=a,则：a4=a＋3d、b4=ad³,得：a＋3d=ad³,则：3d=a(d³－1)-----------------------------(1)同理,

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！

设an=a1+(n-1)d,bn=b1d^(n-1)=a1d^(n-1)a3=a1+2d,b3=a1d^2a5=a1+4d,b5=a1d^4a1+2d=5a1d^2a1+4d=5a1d^4d^2=1+