Question

在等差数列{an}中，a3+a4+a5=84，a9=73 求数列{an}的通项公式

Answer 1

an=a1+(n-1)d
a3+a4+a5=84
3a1+9d=84 (1)
a9=73
a1+8d=73 (2)
3(2)-(1)
15d=135
d=9
a1=1
an = 1+ (n-1)9 = 9n-8
(2)
m=1
9n-8 > 9
n=2
9n-8 < 92
9n <100
n=11
b1= 10 ; n=2,3,...,11
m=2
9n-8 > 18
n=3
9n-8 < 184
n=21
b2= 19 ; n=3,4,...,21
m=3
9n-8 > 27
n=4
9n-8 <273
n=31
b3= 28 ; n=4,5,...,31
bm = 10m - m+1 = 9m+1
Sm = m(11+9m)/2

Answer 2

3a4=84 求出a4=28
a9-a4=a1+8d-a1-3d=5d=45 d=9
通项公式：an=a1+（n-1）d=9n-8

Answer 3

等差数列{an}中，a3+a4+a5=84>>>>>>>>2*a4+a4=84>>>>>>>a4=28
a5=73>>>>>>>>>>>数列{an}的公差d=45

a3=-17
a5-a3=a3-a1>>>>>>>>a1=2*a3-a5=-73-34=-107
等差数列{an}的通项公式an=a1+(n-1)d
=-107+45*（n-1）=45n-152

n∈N﹡