已知树列(an)是公差为2的等差数列,且a1,a2,a5成等比数列,a2为多少?
问题描述：
已知树列(an)是公差为2的等差数列,且a1,a2,a5成等比数列,a2为多少?
答
设a2为x，则a1=x-2，a5=x+6.
a2*a2=a1*a5，即x^2=(x-2)(x+6)
解方程得a2=x=3
答
a2=a1+2
a5=a1+4d=a1+2*4=a1+8
a2^2=a1*a5
(a1+2)^2=a1(a1+8)
a1^2+4a1+4=a1^2+8a1
4a1=4
a1=1
a2=3
a5=9
答
3
答
a2=a1±2
a5=a1±2*4=a1±8
a2^2=a1*a5
(a1±2)^2=a1(a1±8)
a1^2±4a1+4=a1^2±8a1
±4a1=4
a1=±1
a2=a1±2=±3
答
a1=a2-d=a2-2
a5=a2+3d=a2+6
a2^2=a1*a5=(a2-2)*(a2+6)
a2=3
答
a1/a2=a2/a5
a5=a1+8
a2=a1+2
所以
a1/a1+2=a1+2/a1+8
a1=1
a2=3