1 強(qiáng)解ak=a1+(k-1)d
an=a1+(n-1)d
ap=a1+(p-1)d
an*an=ak*ap
于是a1^2+2(n-1)*a1*d+(n-1)^2d^2=(a1+(k-1)d)(a1+(p-1)d)
2(n-1)*a1*d+(n-1)^2d^2=a1d((k-1)+(p-1))+(k-1)(p-1)d^2
a1((k-1)+(p-1)-2(n-1))=((n-1)^2-(k-1)(p-1))d
a1=(n^2-2n-kp+k+p)/(k+p-2n)*d=(n^2-kp)/(k+p-2n)*d+d
an/ak=(a1+(n-1)d)/(a1+(k-1)d)=(nk+np-n^2-kp)/(n^2+k^2-2nk)=(p-n)/(n-k)=(n-p)/(k-n)
但選擇題可巧解如樓上所述舉個(gè)特例就可以
2 1+q>q^2 q>0
q^2-q-1