Question

等差数列an前N项和为Sn,a7=15,S13=？

详细过程

Answer 1

从等差数列的基本性质可知：
a1+a13=a2+a12=a3+a11=a4+a10= a5+a9=a6+a8=2a7
S13=（a1+a13）+（a2+a12）+（a3+a11）+（a4+a10）+（a5+a9）+（a6+a8）+a7
=2a7+2a7+2a7+2a7+2a7+2a7+a7
=13a7
s13=13a7=13*15=195

Answer 2

s13=(a1+a13)*13/2
=2a7*13/2
=13a7
=13*15
=195

Answer 3

s13=（a1/2+a13/2 ）乘以13=13a7=195 拓展s2n-1=（2n-1）an 记住好用。

Answer 4

a7=a1+6d
S13=13(a1+a13)/2=13(a1+a1+12d)/2=13(a1+d)=13a7=195

Answer 5

S13=13(a1+a13)/2
a1+a13=2a1+12d=2a7
S13=13(a7+a7)/2=13×15=195