设{an}是等差数列,{bn}是各项都为正数的等比数列,且a1=b1=1,a3+b5=21,a5+b3=13,求{an},{bn}的通项公式.已知数列{an},那么“对任意n∈N,点pn(n,an)都在直线y=2x+1上”是“{an}为等差数列”的( )A.必要不充
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![设{an}是等差数列,{bn}是各项都为正数的等比数列,且a1=b1=1,a3+b5=21,a5+b3=13,求{an},{bn}的通项公式.已知数列{an},那么“对任意n∈N,点pn(n,an)都在直线y=2x+1上”是“{an}为等差数列”的( )A.必要不充](/uploads/image/z/11439918-54-8.jpg?t=%E8%AE%BE%7Ban%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%7Bbn%7D%E6%98%AF%E5%90%84%E9%A1%B9%E9%83%BD%E4%B8%BA%E6%AD%A3%E6%95%B0%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E4%B8%94a1%3Db1%3D1%2Ca3%2Bb5%3D21%2Ca5%2Bb3%3D13%2C%E6%B1%82%7Ban%7D%2C%7Bbn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F.%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%2C%E9%82%A3%E4%B9%88%E2%80%9C%E5%AF%B9%E4%BB%BB%E6%84%8Fn%E2%88%88N%2C%E7%82%B9pn%28n%2Can%29%E9%83%BD%E5%9C%A8%E7%9B%B4%E7%BA%BFy%3D2x%2B1%E4%B8%8A%E2%80%9D%E6%98%AF%E2%80%9C%7Ban%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E2%80%9D%E7%9A%84%EF%BC%88+%EF%BC%89A.%E5%BF%85%E8%A6%81%E4%B8%8D%E5%85%85)
设{an}是等差数列,{bn}是各项都为正数的等比数列,且a1=b1=1,a3+b5=21,a5+b3=13,求{an},{bn}的通项公式.已知数列{an},那么“对任意n∈N,点pn(n,an)都在直线y=2x+1上”是“{an}为等差数列”的( )A.必要不充
设{an}是等差数列,{bn}是各项都为正数的等比数列,且a1=b1=1,a3+b5=21,a5+b3=13,求{an},{bn}的通项公式.
已知数列{an},那么“对任意n∈N,点pn(n,an)都在直线y=2x+1上”是“{an}为等差数列”的( )
A.必要不充分条件 B.充分不必要条件
C.充要条件 D.既不充分又不必要条件
设等差数列{an}公差d≠0,a1=9d,若ak是a1与a2k的等比中项,则k等于( )
A.2 B.4 C.6 D.8
设{an}是等差数列,{bn}是各项都为正数的等比数列,且a1=b1=1,a3+b5=21,a5+b3=13,求{an},{bn}的通项公式.已知数列{an},那么“对任意n∈N,点pn(n,an)都在直线y=2x+1上”是“{an}为等差数列”的( )A.必要不充
(1)
因为a3+b5=21,a5+b3=13,{an}是等差数列,{bn}是等比数列
所以a1+2d+b1*q^4=21,a1+4d+b1*q^2=13
因为a1=b1=1
所以2d+q^4=20,4d+q^2=12
2d+q^4=20方程乘以2得4d+2*q^4=40
用4d+2*q^4=40减去4d+q^2=12得2*q^4-q^2-28=0即(2*q^2+7)*(q^2-4)=0
所以2*q^2=-7或q^2=4
当2*q^2=-7时q^2=-3.5(不符合,舍去)
当q^2=4时q=2或-2
因为bn}是各项都为正数的等比数列
所以q=2
综上所述得q=2
带入4d+q^2得d=2
所以 an=2n-1
bn=2^(n-1)
(2) B.充分不必要条件
∵点Pn(n,an)都在直线y=2x+1上
∴an=2n+1,
∴“{an}为等差数列,
若“{an}为等差数列,可设an=2n+2,则点Pn(n,an)都不在直线y=2x+1上,
∴对任意的n∈N*,点Pn(n,an)都在直线y=2x+1上”是“{an}为等差数列”的充分而不必要条件,
故选B.
(3) a1=9d
ak=a1+(k-1)d=(k+8)d
a2k=a1+(2k-1)d=(2k+8)d
ak是a1与a2k的等比中项,
ak^2=a1*a2k
(k+8)^2d^2=9d*(2k+8)d d不为0
k^2+16k+64=18k+72
k^2-2k-8=0 k=4或k=-2 k为正整数
k=4