已知{an}是等差数列，前n项和记为Sn，a3=7,s4=24.求通项公式

已知{an}是等差数列，前n项和记为Sn，a3=7,s4=24.求通项公式

因为是等差数列
a3=a1＋2d=7
S4=4a1＋4(4-1)/2d=24
两式解得a1=3,d=2
an=a1＋(n-1)d
=3＋(n-1)2
=2n＋1

已知数列{an}是等差数列,其前n项和为sn,a3=7,s4=24 ,sn=na1+n(n-1)d,4a1+4(4-1)d=24, a1+2d=7,d=-1,a1=9,设p、q是正整数,且p≠q,s(p+q)=9(p+q)-(p+q)(p+q-1)=10(p+q)-(p+q)², (s2p+s2q)/2 =[18p-2p(2p-1)+18q-2q(2q-1)]/2=10(p+q)-2(p²+q²),p²+q²>2pq, 2(p²+q²)>(p+q)²,-2(p²+q²)<-(p+q)²,10(p+q)-2(p²+q²)<10(p+q)-(p+q)², s(p+q)>1/2(s2p+s2q) .