设{an}为递增等差数列,Sn为其前n项和,满足a1*a2-a5=S10,S11=33.

题目

设{an}为递增等差数列,Sn为其前n项和,满足a1*a2-a5=S10,S11=33.
(1)求数列{an}的通项公式及前n项的和Sn;
(2)试求所有的正整数m,使a(m+1)*a(m+3)/a(m+2)为数列{an}中的项.

答案

S11=33 11a1+11*10/2*d=33 a1+5d=3
a1a3-a5=S10 a1(a1+2d)-a1-4d=10a1+10*9/2*d (3-5d)(3-3d)-3+2d=30-50d+45d=30-5d
9-24d+15d^2-3+2d-30+5d=15d^2-4d-24=0 d>0
d=2
a1=-7
an=-7+2(n-1)=2n-9
sn=(-7+2n-9)n/2=n(n-8)=n^2-8n
a(m+1)a(m+3)）/am+2=(2m-7)(2m-3)/(2m-5)=(4m^2-20m+21)/(2m-5)=(4m^2-2m-24)/(2m-5)-9=2m+4-4/(2m-5)-9=2x-9
4-4/(2m-5)=0
m=3