1.已知是an等比数列,(a4+a8)(a6+a10)=49,则a5+a9等于.2.已知等差数列an的公差是d,前n项和Sn=-n^2,则通项公式是.

Question

1.已知是an等比数列,(a4+a8)(a6+a10)=49,则a5+a9等于_________.
2.已知等差数列an的公差是d,前n项和Sn=-n^2,则通项公式是_________.

Answer 1

第一题：
a5+a9 = 7或者-7

因 a6 = a4*q^2, a10 = a8*q^2
则
(a4+a8)(a6+a10)=(a4+a8)(a4*q^2+a8*q^2) = [(a4+a8)*q] = 49
所以 (a4+a8)*q = +/- 7
所以 a5+a9 = (a4+a8)*q = +/- 7
第二题：
a1 = S1=-1
an = Sn - S(n-1) = -n^2 - [-(n-1)^2] = -2n+1
易知n=1时满足.
所以通项公式为 an = -2n+1