In the proportional sequence AN, a1 2 If the sequence An 1 is also a proportional series, then a3

According to an-1 is a proportional series, it can be obtained.

a3-1) (a2-1) = (a2-1) (a1-1)a1q 2-1) (a1q-1) = (a1q-1) (a1-1) Substitute the value of a1 into the above equation, q 2-2q+1=0, q=1a3=a1q 2=2

qan=2*q^(n-1)

an-1=2*q^(n-1)-1

an-1 is a proportional series.

a2-1)/(a1-1)=2q-1

a3-1)/(a2-1)=(2q^2-1)/(2q-1)2q-1=(2q^2-1)/(2q-1)

Solve q and substitute q into a3=2*q 2.

a1=1a2-a1=a1*1/3=1/3

an-a(n-1)=a1*(1 3) (n-1)=1 3 (n-1) add the left and right sides respectively:

Left = a1+a2-a1+.an-a(n-1)=an1+1/3+..1/3^(n-1)

1*(1-1/3^n)/(1-1/3)

an=(3/2)*(1-1/3^n)=3/2-3/(2*3^n)sn=3n/2-(3/2)*(1/3+1/3^2+..1/3^n)3n/2-(3/2)*(1/3)*(1-1/3^n)/(1-1/3)3n/2-(3/4)*(1-1/3^n)

an=2a(n+1)*a(n-1)=an 2[a(n+1)+1][a(n-1)+1]=(an+1) 2 times open That is, the slag has an 2+a(n+1)+a(n-1)+1=an 2+2an+1a(n+1)+a(n-1)=2an=2[a(n+1)+a(n-1)] The only thing that can be said is that the macro number column is a constant number series an=2 .

Q is 2 and the second question is 64

The third question is 62, 9, 1 2

62,1, in the known proportional series, a1=2, a2=4 find the common ratio q of (1) the proportional series

2) The sixth term of the proportional series, A6

3) The sum of the first five terms of the proportional series, s5

Summary. Let the common ratio of the proportional series be r, then we have: a1 + a2 = 3a3 - a1 = 3 express a3 as the formula a1 and r, and get:

a3 = a1 * r 2 substituting the above equation into the second equation, and the solution is: a1 * r 2 - a1 = 3 simplification yields: a1 * r 2 - 1) = 3 and because a4 is the fourth term of the proportional series, there is:

a4 = a1 * r 3 and replace the previously calculated a1 into it, yielding: a4 = 3 r 2 - 1)) R 3 is simplified to yield: a4 = 3 * r) r - 1) 2 At this point, we get the general formula for the proportional series.

However, since the question does not give other conditions, the value of the common ratio r cannot be determined, so the specific value of A4 cannot be calculated.

In the proportional series AN, a1+a2=a3-a1=3, a4=hello, dear, good evening.

Yes. Dear, take a look, according to your own understanding, the answer is as follows.

Let the common ratio of the proportional hand series be r, then there is: a1 + a2 = 3a3 - a1 = 3 express a3 as the equation a1 and r, and get: a3 = a1 * r 2 substitute the above equation into the second equation and solve:

a1 * r 2 - a1 = 3 simplification can be obtained Bi Liangsheng: a1 * r 2 - 1) =3 and because a4 is the fourth term of the proportional series, there is: a4 = a1 * r 3 replace the a1 and obtained earlier, and get:

a4 = 3 r 2 - 1)) R 3 is simplified to obtain: A4 = 3 * R) R - 1) 2 At this point, we get the general formula for the number of proportional key pants. However, since the question does not give other conditions, the value of the common ratio r cannot be determined, so the specific value of A4 cannot be calculated.

Dear, I would like to consult if there are no other conditions.

Not. Then I'm thinking about it.

Kiss how much and wait a little longer.

Dear, this has an original question.

I feel like this question is wrong.

It's finally solved.

Choose b pro, I'm sorry for waiting so long.

There is a process. Yes, there is.

Let the common ratio of the proportional series be r, and the first term is a1, then there is: a1 + a2 = 3a3 - a1 = 3 The formula that expresses a2 and a3 as a1 and r: a2 = a1 * ra3 = a2 * r = a1 * r 2 into the first two equations to repent

a1 * 1 + r) =3a1 * r 2 - 1) =3 to solve the values of a1 and buried r: a1 = 1r = 2 Therefore, the general equation for this equal ratio series is an = 2 (n-1), then a4 = 8.

Take a look, dear.

How is this equation solved?

Compare the two equations.

Can you write a little more about how to solve this equation?

Take a look for yourself.

1`.Establish.

The common ratio q, then a1 2q=-1 3, a1q 2=1 9

a1a2<0, a3>0, then a1>0, thus q<0

Multiply the two formulas. a1^3q^3=-

a1q=a2=-1 3, so a1=1, q=-1 3

an=(-1/3）^(n-1)2.

bn=(n+1)[1-1/2+1/2-1/3+．．1/n-1/(n+1)]=(n+1)*(1-1/(n+1)=(n+1)*n/(n+1)=n

bn/an=n*(-3)^(n-1)

The following can be solved by applying the dislocation subtraction method.

bn is solved using the method of fission elimination).

an=2q^(n-1)

Because it is an equal proportion of the holding of the Liang Shi Zhatong series, the segment limb:

an+1) 2=[a(n-1)+1][a(n+1)+1] i.e.: an 2+2an+1=a(n-1)a(n+1)+a(n-1)+a(n+1)+1, and an 2=a(n-1)a(n+1).

So: 2an=a(n-1)+a(n+1).

That is: 4q (n-1)=2q (n-2)+2q n that is: q 2-2q+1=0

q=1an=2

a1a2a3=8, that is, a1 3 q 3=8, a1 q=2, a1=2 qa1+a2+a3= -3, that is, a1 (1+q+q 2)= 3, substitute a1=2 q, and you can get it.

2+2q+2q^2= -3q，2q^2+5q+2=0，(2q+1)(q+2)=0

q= -1/2，a1= -4

or q = -2, a1 = -1

an=(-4)×(1/2)^(n-1)

an= -2)^(n-1)

Since a1a3=a2 squared, the cube of a2 from a1a2a3=8 is equal to 8, so a2=2, thus a1=-1 and q=-2, or a1=-4 and q=- so an=-(2) (n-1) or an=-4(- n-1).

Because a1a2a3=8, q less than 0 is not valid and should be rounded.