a13=(a1+a25)/2
=(a1+a25+a3+a23)/4
=2(a26)/4
=(2x12)/4
=6
a1,a3之等差中項為a2,(a1+a3)/2=a2,得到a2=3/2
a23,a25之等差中項為a24,(a23+a25)/2=a24,得到a24=21/2
公差以d表示
a2+22d=a24,得到d=9/22
a13=a2+(11*d),得到a13=6
答:a13=( 6 )
a1+a3=a1+(a1+2d)=2a1+2d=3
a23+a25=(a1+22d)+(a1+24d)=2a1+46d=21
兩式相減
得: 44d=18,d=18/44,a1=12/11
a13=a1+12d=66/11=6
a1+a1+2d=3
->>a1+d=1.5
a1+22d+a1+24d=21
->>a1+23d=10.5
22d=9
11d=4.5
a13=a1+12d=a1+d+11d=1.5+4.5=6
ans:6
a=12/11,d=9/11
a13= 12/11+12*9/11=120/11