若an为等差数列,a7等于4,则s13等于多少
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a1+a7+a13=3a7=3、a7=1.S13=13a7=13.S13+ak=14,则ak=1.若公差d=0,则k为任意正整数.若公差d不为0,则k=7.
由等差数列的性质可得a3+a7=2a5=-6,解得a5=-3.又a1=-11,设公差为d,所以,a5=a1+4d=-11+4d=-3,解得d=2.则an=-11+2(n-1)=2n-13,所以Sn=n
设a4=m,公比为q,所以a6=mq2,a7=mq3a4+a7=2a6m+mq3=2mq21+q3=2q2(q-1)(q2-q-1)=0∵q≠1∴q2-q-1=0∴q=1+52或1−52(舍)∴a4+
根据等差中项的性质得a3+a7=2a5a4+a6=2a5所以a3+a4+a5+a6+a7=(a3+a7)+(a4+a6)+a5=5a5=450得a5=90所以a2+a8=2a5=2×90=180选C
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由题知a4+a6=2(a5+a7)=2(a4q+a6q)=2q(a4+a6),由a4+a6≠0得q=12.故答案为12
汗总得给个a1是多少吧
因为1/(an+1)为等差数列,设bn=1/(an+1)b3=1/3b7=1/2,b7=b3+4dd=(b7-b3)/4公差d=(1/2-1/3)/4=1/24b8=b7+d=1/2+1/24=12/
an是等差数列那么a3+a11=2a7,由2a3-a7^2+2a11=0;则4a7-a7^2=0所以a7=0或等于a7=4;由于an每项不为0,所以a7=4;那么b6*b8=b7^2=a7^2=16
因为{an}是公差为-2的等差数列,∴a3+a6+a9++a99=(a1+2d)+(a4+2d)+(a7+2d)+…+(a97+2d)=a1+a4+a7++a97+33×2d=50-132=-82.故
a5+a6+a7+a8=242(a1+a12)=24a1+a12=12
a3=2,a7=1,若{1/(an+1)}为等差数列公差d=(1/a7+1-1/a3+1)/(7-3)=1/241/(a11+1)=1/3+1/24*(11-3)=2/3a11=1/2
设an=a1+(n-1)d(1):a6=a1+5d>0.(1)a7=a1+6d
a4=a1+3d=7a7=a1+6d=1a1=13d=-2an=a1+(n-1)d=13-2(n-1)=-2n+15
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*
a4+3d=a77+3d=4d=-1a15=a7+8d=4-8=-4(d是公差)
解法1:∵{an}为等差数列,设首项为a1,公差为d,∴a2+a3+a10+a11=a1+d+a1+2d+a1+9d+a1+10d=4a1+22d=48;∴2a1+11d=24;∴a6+a7=a1+5
a5=a4*qa7=a4*q^3a6=a4*q^22(a5+a7)=a4+a62(a4*q+a4*q^3)=a4+a4*q^2a4不等于0两边同时÷a42q+2q^3=1+q^22q(1+q^2)=1
a2=a1+da3=a1+2da7=a1+6da3^2=a2*a7(a1+2d)^2=(a1+d)(a1+6d)a1^2+4a1d+4d^2=a1^2+7a1d+6d^23a1d+2d^2=0d≠0∴
a1+a4+a7+···+a25=9a1+a25=2a13a4+a22=2a13a7+a19=2a13a10+a16=2a13所以a1+a4+a7+···+a25=9a13=9a13=1a14=1-4