等差数列{an}已知a3＋a8＝24,求S10

来源：学生作业帮助网编辑：六六作业网时间：2024/12/25 10:54:03

等差数列{an}已知a3＋a8＝24,求S10等差数列{an}已知a3＋a8＝24,求S10等差数列{an}已知a3＋a8＝24,求S10a3=a1+2d,a8=a1+7d,so:a3+a8=2a1+

等差数列{an}已知a3＋a8＝24,求S10
等差数列{an}已知a3＋a8＝24,求S10

等差数列{an}已知a3＋a8＝24,求S10
a3=a1+2d, a8=a1+7d,
so: a3+a8=2a1+9d=24
cause:Sn=a1*n+n(n-1)d/2
then: s10=a1*10+10(10-1)d/2
=10a1+45d=5(2a1+9d)=5*24=120
the answer is 120

120

an=a1+(n-1)a
a3=a1+2a
a8=a1+7a
所以a3+a8=a1+2a+a1+7a=2a1+9a=24
sn=na1+((n(n-1))/2)a
s10=10a1+45a
=5(2a1+9a)
=5*24
=120

∵{an}为等差数列
∴a3＋a8＝24=a1+a10
∴S10=(a1+a10)×10/2
=24×5
=120

等差数列{an}已知a3＋a8＝24,求S10 在等差数列an中,已知a2＋a3+a10+a11=36,求a5+a8 已知等差数列{an},a3=22,a8=7,求|a1|+|a2|+.+|an|（要过程）已知等差数列｛an｝中,a3+a6＝17,a1·a8＝-38且a1＜a8 求an的通项公式已知等差数列｛An｝的前n项和为Sn,A3＝6,A8＝16,求S10的值已知等差数列{An} A6=5 A3+A8=5 求 a9=? 在等差数列{an}中,已知a6=5,a3+a8=5,求a1与d 已知{an}为等差数列,a3+a8=22,a6=7,求a5 等差数列an　,已知a2+a3+a8+a11=48,求a 6+a 7 已知等差数列{an}中,a3+a11=10求a6+a7+a8 已知（an)为等差数列,a3+a4+a8=9.求S9 已知等差数列{an}且a8=16,a1+a2+a3=12求等差数列{an}通向公式已知等差数列｛an｝中,a3+a5=18,a4+a8=24,则公差已知数列{An}为等差数列,A4A7=-512,A3+A8=124,公差大于0,求AN 已知等差数列｛an｝满足a3+a6=-1/3,a1*a8=-4/3且a1＞a8求数列｛an}通项公式谢谢了, 等差数列an中a3+a8>0 s9 在等差数列{an}中,已知a3=7 a5=11 求a8在等差数列{an}中,已知a3=7 a5=11 求a8 想半天了 2011╮緈鍢給 01-02 10:41:43已知等差数列｛An｝中,a3＋a6＝17,a1a8＝负38且a1小于a8,求｛An｝的通项公式