作业帮公差d不等于0,a5=10
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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
a1=a3-2d=8-2d,a5=a3+2d=8+2d,a17=a3+14d=8+14da1,a5,a17成等比数列所以(8+2d)^2=(8-2d)(8+14d)即d^2-2d=0因为d≠0所以d=
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,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
因为a5=a1+4d,a9=a1+8d,a15=a1+14d且a5a9a15成等比数列所以(a1+8d)^2=(a1+4d)(a1+14d)即(a1)^2+16a1*d+64d^2=(a1)^2+18
a1+4d＝a5＝0a1+9d＝a10＝10两式相减得5d＝10.所以公差d＝2,导入得a1＝－8
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
a5=a1+4d；a10=a1+9d；两式相减5d=10；解得d=2；带入任意一式,解得a1=-8；
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
∵a4+a5+a6=15∴a5=5∴a4-1=a5-d-1=4-1a5+1=6,a6+6=a5+d+6=11+d∵a4-1,a5+1,a6+6成等差数列∴（a5+1）/（a4-1）=（a6+6）/（a
由题知a1*a9=(a3)^2设公差为d那么a1*(a1+8d)=(a1+2d)^2即(a1)^2+8a1d=(a1)^2+4a1d+4d^2整理得4a1d=4d^2由d不等于0知a1=d所以a1+a
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+
a2+a3+a5=34a2+a2+d+a2+3d=343a₂+4d=34d=(34-3a₂)/4a2×a5=52a2×(a2+3d)=52a₂²+3a&#
a1+4d=8(a1+8)*5/2=10解得a1=-4d=3
公差d=(a12-a5)/7=3,a1=a5-4d=10-4*3=-2