Question

已知等差数列{an}首项与公差相等{an}的前几项和记作sn且s20=840求a及通项公式？求数列

已知等差数列{an}首项与公差相等{an}的前几项和记作sn且s20=840求a及通项公式？求数列前多少项和为84？

Answer 1

a1=d

an=a1+(n-1)d=nd

sn=na1+n(n-1)d/2=(n+1)nd/2

s20=(20+1)*20*d/2=210d=840

d=4

a1=4

an=4n

 

2.

sn=(n+1)nd/2=2(n+1)n=84

(n+1)n=42

n=6

Answer 2

an = nd
S20=840
210d =840
d= 4
an = 4n

Sn = 84
(4n+4)n/2 = 84
n^2+n-42=0
(n-7)(n+6)=0
n=7
S7=84

Answer 3

a1=d
an=a1+(n-1)d=nd
sn=na1+(n-1)nd/2=(n+1)nd/2
s20=(20+1)*20*d/2=840 d=4
an=4n
sn=2n(n+1)=84 n=6