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已知等差数列{An}的公差D≠0,且A1,A3,A9成等比数...

(Ⅰ)由题意知a32=a1a9,即 (2+2 d)2=2×(2+8d)…(3分)得:d2-2d=0,解得d=2或d=0(舍)∴an=2n.…(5分)(Ⅱ)数列{2an-1}的通项为2an-1=22n-1=4n-1,…(7分)∴Sn=41+42+43+…+4n-n=43×(4n-1)-n. …(10分)

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=3d 所以 (a1+a3+a9)/(a2+a4+a10) =[(a3-2d)+a3+(a3+6d)]/[(a3-d)+(a3+d)+(a3+7d)] =[3(a3)+4d]/[3(a3)+7d] ...

a1,a3,a9成等比数列 有a3*a3=a1*a9,即 (a1+2d)^2=a1*(a1+8d) 括号打开化简的4d*d=4a1*d 有a1=d (a1+a3+a9)/(a2+a4+a10)=(3a1+10d)/(3a1+13d)=13/16 谢谢,祝你学习进步!

(1)设等差数列的公差为d,则a3=1+2d,a9=1+8d.因为a1、a3、a9成等比数列,所以(1+2d)2=1?(1+8d),解得:d=1.所以an=n;(2)a1+a4+a7+…+a40=1+4+7+…+40=14×1+14×(14?1)2×3=287.

设等差数列的公差为d,首项为a1,所以a3=a1+2d,a9=a1+8d.因为a1、a3、a9成等比数列,所以(a1+2d)2=a1(a1+8d),解得:a1=d.所以公比q=a3a1=a1+2da1=3dd=3,故选:D.

∵等差数列{an}的公差和首项都不等于0,且a2,a4,a8成等比数列,∴a42=a2a8,∴(a1+3d)2=(a1+d)(a1+7d),∴d2=a1d,∵d≠0,∴d=a1,∴a1+a5+a9a2+a3=15a15a1=3.故选:B.

设该数列的公差为d,前n项和为Sn,则∵a1+a3=8,且a4为a2和a9的等比中项,∴2a1+2d=8,(a1+3d)2=(a1+d)(a1+8d)解得a1=4,d=0或a1=1,d=3∴前n项和为Sn=4n或Sn=3n2?n2.

解:充分利用等比数列、等差数列的性质。 a6^2=a3*a9 →(a3+a9)^2=a3^2+a9^2+2*a3*a9=a3^2+a9^2+2*a6^2 …① 2*b7=b4+b10→(b4+b10)^2=4*b7^2=4*a6^2 ...② ①-②得:(a3+a9)^2-(b4+b10)^2=a3^2+a9^2-2*a6^2=(a3-a9)^2≥0 所以:(a3+a9)^2≥(b4+b10)^2 |a3+...

a1=a1 a3=a1+2d a9=a1+8d a3*a3=a1*a9 (a1+2d)^2=a1*(a1+8d) a1^2+4d^2+4a1*d=a1^2+8a1*d ==> a1=d (a1+a3+a9)/(a2+a4+a10) =[a1+ (a1+2d)+(a1+8d)] / [(a1+d)+(a1+3d)+(a1+9d)] =[3a1+10d]/[3a1+13d] =[13d]/[16d] =13/16

设公差为d, a3=a1+2d,a9=a1+8d a3²=(a1+2d)²=a1²+4a1d+4d² a1a9=a1²+8a1d 4a1d+4d²=8a1d d不为0,所以a1=d (a1+a3+a9)/(a2+a4+a10)=(a1+3a1+9a1)/(2a1+4a1+10a1)=13/16

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