等差数列{an}已知a3＋a8＝24,求S10
问题描述：
等差数列{an}已知a3＋a8＝24,求S10
答
∵{an}为等差数列
∴a3＋a8＝24=a1+a10
∴S10=(a1+a10)×10/2
=24×5
=120
答
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
答
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
答
120