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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)

等差数列{an}中,a1=a1,a3=a1+2d,a9=a1+8d,因为a1、a3、a9恰好是某等比数列,所以有a32=a1a9,即(a1+2d)2=a1(a1+8d),解得d=a1,所以该等差数列的通项为an=nd则a1+a3+a9a2+a4+ a10的值为1+3+92+4+10=1316.故选C.

∵a1,a3,a9成等比数列,∴(a1+2d)2=a1?(a1+8d),∴a1=d,∴a1+a3+a9a2+a4+a10=1316,故答案是:1316.

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=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 若满意请采纳!!谢谢

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!

解:由题意可知,则 a3²=a1×a9,即 (a1+2d)²=a1(a1+8d) a1²+4d²+4a1d=a1²+8a1d 4a1d=4d² 解得:a1=d 故,an=a1+(n-1)d=d+nd-d=nd 答题不易,望采纳~~

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解题如下,一定给个采纳哟。 ∵an为等差数列,a1,a3,a9成等比数列, a1*a9=(a3)^2, a9=a1+8d, a3=a1+2d ∴a1(a1+8d)=(a1+2d)^2 a1^2+8d*a1=a1^2+4d*a1+4d^2 d≠0 ∴d=a1 a1+a3+a9/a2+a4+a10=(a1+a1+2d+a1+8d)/(a1+d+a1+3d+a1+9d) =13a1/16a1=13/16

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=13/16

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