早教吧作业答案频道 -->数学-->
等差数列{An}满足A10=15,求S19
题目详情
等差数列{An}满足A10=15,求S19
▼优质解答
答案和解析
s19=(a1+a19)*19/2
a1+a19=2a10=30
s19=285
a1+a19=2a10=30
s19=285
看了等差数列{An}满足A10=1...的网友还看了以下:
正数列{an}和{bn}满足对任意自然数n,an,bn,an+1成等差数列,bn,an+1,bn+ 2020-04-06 …
已知数列{an}满足a1=1an+1=Sn+(n+1)(1)用an表示an+1(2)证明数列{an 2020-04-27 …
设数列{an}和{bn}满足a1=b1=6,a2=b2=4,a3=b3=3,且数列{an+1-an 2020-05-13 …
已知数列{an}的前n项和为Sn,且满足Sn+n=2an(n∈N*).(1)证明:数列{an+1} 2020-05-13 …
若数列an满足an+1平方-an2=d.其中d为常数,则称数列an为等方差数列,已知等方差数列a满 2020-06-10 …
1)已知数列{an}满足a1=1,n≥2时,an-1-an=2an-1an,求通项公式an2)已知 2020-06-11 …
下面一道有趣的数列大题,大家有空看下吧:数列{an}恒满足等式a(n+1)=1/2an+√3/2× 2020-07-23 …
已知公差不为0的等差数列an满足a2=3,a1、a3、a7成等比数列.(1)求an通项公式(2)数 2020-07-30 …
已知等比数列an的各项均为正数,且满足2a1+a2=8,a2a6=4a5^2已知{an}是一个公差大 2020-10-31 …
(理)数列,若对任意的,满足,是常数且不相等),则称数列为“跳跃等比数列”,则下列关于“跳跃等比数列 2020-11-20 …