Question

高一数学等差数列，急！

计算
(1){an}为等差数列且a1=20,an=54,sn=999,求d及n
{1)已知等比数列a1=-1，a4=64a求q及s
{2)已知等比数列a1=-1，a4=64，求q与s4 回答这个第二题，上面的第二题是错的

Answer 1

1. an =a1+(n-1)d= 20+(n-1)d=54
(n-1)d=34
Sn =na1+ nd(n-1)/2 =20n+34n/2=37n=999
n = 27
d = 34/(27-1) = 17/13
2. a4 = a1*q^3
64 = -q^3
q = -4
s4 = a1(1-q^4)/(1-q) = [(-4)^4 -1]/5 = 51

Answer 2

(1) an = a1 + (n-1) *d = 20 + (n-1)*d = 54
sn = n*a1 + (n-1)*n/2 *d = 999
n = 27, d = 17/13
(2) a4 = a1 *q^3 = -q^3 = 64, q = -4
sn = -1*(1-(-4)^n) / (1-(-4)) = [(-4)^n -1] / 5

Answer 3

Sn=n(a1+a2)/2=n(20+54)/2=999 则n=27
an=a1+(n-1)d=20+(27-1)d=54
d=17/13
第二题看不清

Answer 4

(1) 设公差为d
an=a1+(n-1)d=20+(n-1)d=54
(n-1)d=34
sn=[2a1+(n-1)d]*n/2=[40+34]*n/2=999
n=27 d=34/26=17/13
(2) 设公比为q
a4=a1q^3=64a1 q=4
sn=a1(1-q^n)/(1-q)=-1*(1-4^n)/(1-4)=(1-4^n)/3