设等差数列{an}的前n项和为Sn，若a1a5a9=15，且1a1a5+1a5a9+1a9a1=35，则S9=．

Question

设等差数列{a_n}的前n项和为S_n，若a₁a₅a₉=15，且 (1)/(a1a5)+ (1)/(a5a9)+ (1)/(a9a1)= (3)/(5)，则S₉=___．

Answer 1

设等差数列{a_n}的公差为d，
∵a₁a₅a₉=15，∴（a₅-4d）a₅（a₅+4d）=15．
∵(1)/(a1a5)+(1)/(a5a9)+(1)/(a9a1)=(3)/(5)，
∴(1)/(4d)((1)/(a1)-(1)/(a5))+(1)/(4d)((1)/(a5)-(1)/(a9))+(1)/(8d)((1)/(a9)-(1)/(a1))=(3)/(5)，
化为(1)/(8d)((1)/(a1)-(1)/(a9))=(3)/(5)，即(1)/(8d)((1)/(a5-4d)-(1)/(a5+4d))=(3)/(5)．
解得a₅=3，
则S₉=(9(a1+a9))/(2)=9a₅=27．
故答案为：27．