在等差数列an中,a4+a5+a6+a7+a8=36,求s11

来源:百度知道 编辑:UC知道 时间:2024/06/19 11:10:16
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79.2
设每个数相差k
则a4+a5+a6+a7+a8
=a1+3k+a1+4k+a1+5k+a1+6k+a1+7k
=5a1+30k
S11=11a1+(1+2+...+11)k
=11a1+66k
所以S11/(a4+a5+a6+a7+a8)
=(11a1+66k)/(5a1+30k)
=11(a1+6k)/5(a1+6k)
=11/5
因为(a4+a5+a6+a7+a8)=36
所以S11=(11/5)*36=79.2

因为a4+a5+a6+a7+a8=36
所以5a1+(3+4+5+6+7)d=36
a1+5d=36/5
s11=11a1+(1+2+...10)d
=11a1+5(1+10)d
=11a1+55d
=11(a1+5d)
=11*36/5
=396/5
=79.2

s11=11×(a1+a11)÷2
a1+a11=2×a6
a4+a5+a6+a7+a8=5×a6=36
s11=79.2

a4+a8=a5+a7=2a6,
所以a4+a5+a6+a7+a8=5a6=36,
a6=6,s11=11(a1+a11)/2=11a6=66.