Question

在等差数列{an}中,若3(a3+a5)+2(a7+a10+a13)=24,则s13=

好像是高考题

Answer 1

设公差是d
那么
a3=a1+2d
a5=a1+4d
a7=a1+6d
a10=a1+9d
a13=a1+12d

代入原来的式子，就会得到
12*a1+72d=24
a1+6d=2
而s13=(a1+a13)*13/2
=(a1+a1+12d)*13/2
=2*(a1+6d)*13/2
=26

Answer 2

设公差为D
3(A3+A5)+2(A7+A10+A13)=3(2A1+6D)+2(3A1+27D)=6(A1+3D)+6(A1+9D)
=6(A4+A10)=6(A1+A13)=24
所以A1+A13=4
S13=(A1+A13)*13/2=4*13/2=26