a1=1.公差d不等于0

a(n)=a+(n-1)d,d不为0.[a(3)]^2=a(1)a(9)=a[a+8d]=[a+2d]^2,0=a^2+8ad-a^2-4ad-4d^2=4ad-4d^2=4d(a-d),0=a-d.

不需要分奇偶数cn=an*bn=(4-3n)*(-2)^(n-1)Sn=(4-3*1)(-2)^0+(4-3*2)(-2)^1+……+[4-3(n-1)](-2)^(n-2)+(4-3n)(-2)^(

1）∵a2=b2∴1+d=1×q∵a4=b4∴1+3d=1×q^3组合成方程组后把d=q-1带入1+3d=q^3q^3-3q+2=0q^3-3q+3-1=0q^3-1-3(q-1)=0(q-1)(q^

额..a1=da2=2dq=a2/a1=2d/d=2.

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=

a2=a1+d,a3=a1+2d.,a6=a1+5d,...,a10=a1+9d,若a1,a3,a6成等比数列,则a3^2=a1*a6,(a1+2d)^2=a1*(a1+5d),得到a1=4d.则(a

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=

a4=a1+3d,a8=a1+7da1,a4,a8成等比数列,a4²=a1×a8(a1+3d)²=a1×(a1+7d)整理得,a1=9d所以an=a1+(n-1)d=(n+8)d﹙

已知公差为d(d不等于0）,a1=1,那么：a2=a1+d=1+d,a5=a1+4d=1+4d,a14=a1+13d=1+13d又a2a5a14依次成等比数列,所以：(a5)²=a2*a14

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*

26/29设公差是d,a1,a5,a17的公比为q那么a1q=a1+4d（a1+4d）q=a1+16d联立解得q=3da1=2d代入得到（a1+a5+a17）/（a2+a6+a18）=26d/29d=

a1=a3-2d,a9=a3+6d因为a1,a3,a9成等比数列,所以有(a3)^2=(a1)*(a9)所以(a3)^2=(a3-2d)(a3+6d)所以3d^2=d*(a3)因为d不等于0所以a3=

an=Sn-Sn-1=2^n-1-{2^(n-1)-1}=2x2^(n-1)-1-2^(n-1)+1=2^(n-1)a1=s1=1,所以an通项公式为2^(n-1)b1=a1=1,b3=1+2d,b9

因为a1,a2,a5成等比数列,根据等比中项公式：a2^2=a1xa5(1+d)^2=1x(1+4d)d^+2d+1=4d+1d^2-2d=0d=0或d=2因为d不等于0,所以d=2

因为1/anan+1=1/an*（an+d）=1/d[1/an-1/（an+d）]=1/d[1/an-1/an+1]所以1/a1a2+1/a2a3+…+1/anan+1=1/d[1/a1-1/a2+1

解析：∵a1=4,a7=4+6d,a10=4+9d∴a7^2=a1*a10,即（4+6d)^2=4(4+9d)∵d≠0∴d=-1/3即a1=4,a7=2,a10=1∴q=a2/a1=1/2∴Sn=4*

A2*A2=A1*A4A2=A1+dA4=A1+d得A1=dA10=10dS10=10(A1+A10)/2=110A1=d=2An=2n

是A1,A3,A4等比数列吧?∵A1,A3,A4等比数列∴(a3)²=(a1)×(a4)(a1＋2d)²=(a1)(a1＋3d)a²₁+4d²+4a

a(1)=b(1)=a,a(n)=a+(n-1)d,b(n)=ad^(n-1).a(4)=a+3d=b(4)=ad^3,3d=a[d^3-1],d不为0,因此,d不为1.a=3d/[d^3-1].a(