Freshman Math! Equally, sophomore math help! Proportional series

According to the known results: sn=a1+a1q+a1q*q+.a1q^n1=80...1)

s2n=a1+a1q+a1q*q+..a1q^2n-1=6560...2)

Put the a1 in equations (1) and (2) to get :

sn=a1(1+q+q^2+q^3+..q^n-1)=80...3)s2n=a1(1+q+q^2+q^3+..q^2n-1)=6560...4)

Divide Eq. (4) by Eq. (3) (using the first n terms of the proportional series and the formula):

1-q^2n)/(1-q^n)=82

Multiply the denominator to get :

1-q^2n=82-82*q^n

Consider q n as x, i.e., x2-82x 81=0 cross multiplication:

x = 81 or x = 1 (rounded) (n is a natural number, q cannot be one) and because the largest term in the first n terms is an=a1*q (n-1)=54, i.e., 54=(a1*q n) q....5)

Bringing q n=81 into equations (3) and (5) yields:

81a1=54q

80(q-a1)=80

Solution: a1=2 q=3

Therefore an=a1*q n-1=2*3 n-1

sn=a1(1-q^n)/(1-q)

s2n=a1(1-q^2n)/(1-q)

So sn s2n = (1-q n) (1-q 2n) = 80 6560 = 1 82

1+q^n=82

q^n=81

Since n is a natural number, then q>1 or q<-1

Then the maximum value is the last term, an=54=a1*q (n-1)=a1*q n q

a1(1-q^n)/(1-q)=80

a1*q^n/q=54

q^n=81

Stand-by-side. a1=2 q=3 n=4

an=2*3^(n-1)

∵a5-a1=15，a4-a2=6

a5+a4-(a1+a2)=21

3a3-a3=21

a3=21/2

For example, a, b, and c are proportional sequences, and b = ac

18=a1*q，8=a1*q^3，a1*q³/a1*q=8/18

q²=4/9

q=±2/3

When q=2 3, a1=18 2 3=27

When q=2 3, a1=18 2 3=27

1. A1, A2, A4, A5 are all represented by A1 and Q. Yes: a1(q4-1)=15,a1q(q2-1)=6, the solution can be found a1,q, and then a3

2,b2=ac

3, divide the two formulas, and q= 2

(a1q 2+a1q 3)=11(a1q*a1q 3) gives the relationship between a1 and q.

a1=（1+q）/（11q^2）

a1*[1-q^(2n)]/(1-q)=11a1*q*[1-q^(2n)]/(1-q^2)

q=1/10

So a1=10

So an=10*(1 10) (n-1).

1/[10^(n-2)]

I don't know if it's right or not.

1. Because the general formula of the proportional series is an=a1*q n, (where q is the common ratio of the proportional series), so a3=a1*q 2, a4=a1*q 3, a6=a1*q 5, so a3-a1=8 is a1(q 2-1)=8

a6-a4 = 216, i.e. a1 (q 5 - q 3) = 216a1 * q 3 (q 2-1) = 216

Divided: q 3 = 27, that is, q = 3, substituting q into the formula can find a1 = 1 so the general term of the series an=q (n-1)a1=3 (n-1)2, when sn=121, because the first n terms of the proportional number example of the Hui Zhongbi front move and the general term of sn Peifan are sn=a1*(1-q n) (1-q)=121

where a1 = 1 and q = 3, so there is (1-3 n) (1-3) = 121 and 1-3 n = -242

3 n = 243, i.e. 3 n = 3 5

So n=5

1) Set the ratio of the first tour of the equal ratio series next to the celery to q, so there is.

a3=a1*q^2

a4=a1*q^3

a6=a1*q^5

then a3-a1=

a1*(q^2-1)=8

a6-a4=a1*q 3*(q 2-1)=216 The following formula divides the above formula by the pin, and Q=3 can be obtained, so A1=1 then an=1*q (n-1).

2) Formulas by proportional sequences.

sn=a1*(1-q^n)/(1-q)=(1-3^n)/(2)=121

3^n=245n=5

(1) Ke1+E2 and E1+Ke2 are collinear, and there is such that Ke1+E2= (E1+Ke2), i.e., (K-)E1+(1- K)E2=0

e1 and e2 are non-zero non-collinear vectors, k- =0 and 1- k=0

k= 1 2) by (ke1+e2) (e1+ke2)=0,k|e1|2+（k2+1）e1•e2+k|e2|2=0, K 22+(K2+1) 2 3 cos60°+K 32=0 4K+3K2+3+9K=0 3K2+13K+3=0, K= (-13 133) 6

The solution consists of a1+a2=2

a1+a2+a3=-6

Know a1 + a1q = 2

a1+a1q+aq^2=-6

The two formulas are divided. (1+q) (1+q+q 2)=-1 3, then 3+3q=-1-q-q 2

Then q 2 + 4 q + 4 = 0

i.e. (q+2) 2=0

The solution yields q=-2

Substitute q=-2 into a1+a1q=2

then a1=-2

then an=a1q n=(-2) n