Question

已知等差数列an的前n项和为Sn，a5＝7，则S9＝？？

Answer 1

在等差数列an中有
a1+a9=2a5
则S9=9(a1+a9)/2
=9*2a5/2
=9a5
=9*7
=63

Answer 2

a5=a1+4d
s9=a1+a2+a3+a4+a5+a6+a7+a8+a9
=(a1+a9)+(a2+a8)+(a3+a7)+(a4+a6)+a5
=2a5+2a5+2a5+2a5+a5
=9a5
=63

Answer 3

等差数列，a1=a5-4d，a2=a5-3d，a3=a5-2d，a4=a5-d，a6=a5+d，a7=a5+2d，a8=a5+3d，a9=a5+4d，则S9=a1+a2+...+a9=9a5=63

Answer 4

S9=9×(a1 + a9)÷2
=(9/2)×(a1 + a1 + 8d)
=(9/2)×(2a1 + 8d)
=(9/2)×[2(a1 + 4d)]
=9×a5
=9×7
=63

Answer 5

S9/S5=[(a1+a9)*9/2]/[(a1+a5)*5/2]
=(9/5)*(a1+a9)/(a1+a5)
=(9/5)*(2*a5)/2*a3)
=(9/5)*(5/9)
=1