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Since a3 2=a1*a9, then (a1+2d) 2=a1*(a1+8d), the relationship between a1 and d is d=a1

Then substitute, i.e., (a1+a1+2d+a1+8d) (a1+d+a1+3d+a1+9d)=(3a1+10d) (3a1+13d), substitute d=a1, eliminate d, and the final result is 13 16

Let the first term of the difference series be A1, and the tolerance is D

Because a1, a3, and a9 are proportional sequences, (a3) 2=a1*a9a1+2d) 2=a1*(a1+8d) gives a1=da1 a3 a9) (a2 a4 a10)=(3a1+10d) (3a1+13d)=13 16

Because a1, a3, a9 are proportional sequences.

So. a1*a9=(a3)^2

That is, a1*(a1+8d)=(a1+2d) 2 is solved to get a1=d

So. a1＋a3＋a9）／（a2＋a4＋a10）=13/16

Because a1, a3, a9 are proportional sequences.

So a1a9=a3 2

a1(a1+8d)=(a1+2d)^2

a1^2+8a1d=a1^2+4a1d+4d^2a1d=d^2

d(a1-d)=0

Because the tolerance of the equal difference series is not equal to zero.

So a1-d=0, a1=d

an=a1+(n-1)d=d+(n-1)d=nda1＋a3＋a9）／（a2＋a4＋a10）

This is a series of equal differences, which can be used using the equation: (the first term is similar to this last term) x terms number 2. Therefore, the answer to Gao Xun's case is: 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 (30 39) x10 2 345

Solution: There are two formulas for summation of proportional sequences, which one are you going to use?

Of course, you can't use them all, you have to choose according to whether the common ratio q is 1 or not.

However, there is no special explanation in the title, so we can only judge whether the common ratio q is 1 according to the meaning of the question

Let's start with the assumptions:

If q=1, then there is.

sn=nq，s(n+1)=(n+1)q，s(n+2)=(n+2)q

s(n+1), sn, and s(n+2) are in a series of equal differences.

s(n+1)+s(n+2) =2sn

n+1)q+(n+2)q=2nq

nq+q+nq+2q=2nq

2nq+3q=2nq

3q=0q=0

A sequence is a proportional series.

The common ratio q≠0 and other proportional sequences cannot contain 0, and the natural common ratio cannot be 0! )

Therefore, the conclusion contradicts the nature of the proportional series, i.e., the assumption is not true, and the common ratio q≠1

sn=a1(1-q^n)/(1-q)，s(n+1)=a1[1-q^(n+1)]/(1-q)，s(n+2)=a1[1-q^(n+2)]/(1-q)

s(n+1), sn, and s(n+2) are in a series of equal differences.

s(n+1)+s(n+2) =2sn

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

a1[1-q^(n+1)]+a1[1-q^(n+2)]=2a1(1-q^n)

A sequence is a proportional series.

The first item is A1≠0

1-q^(n+1)]+1-q^(n+2)]=2(1-q^n)

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

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

q^(n+1)+q^(n+2)=2*q^n

q^(n+1)+q^(n+2)-2*q^n=0

q^n*q+q^n*q^2-2*q^n=0

q^n(q^2+q-2)=0

q^n(q-1)(q+2)=0

q≠0q^n≠0

q-1)(q+2)=0

q≠1q-1≠0

q+2=0q=-2

To sum up, the value of the common ratio q is -2

If it's a solution question, you should do it, but since this question is a fill-in-the-blank question, the method on the second floor is simple and time-saving.

How's that, landlord, satisfied?

Let the first term be a, a≠0

If q=1sn=na

s(n+1)=(n+1)a

s(n+2)=(n+2)a

2na=a(n+1)+a(n+2)

2an=a(2n+3)

2an=2an+3a

a=0, contradictory.

So q≠1sn=a(1-q n) (1-q).

s(n+1)=a[1-q^(n+1)]/(1-q)s(n+2)=a[1-q^(n+2)]/(1-q)2a(1-q^n)/(1-q)=a[1-q^(n+1)]/(1-q)+a[1-q^(n+2)]/(1-q)

2a(1-q^n)=a[1-q^(n+1)]+a[1-q^(n+2)]

2a-2aq^n=a-aq^(n+1)+a-aq^(n+2)-2aq^n=-aq^(n+1)-aq^(n+2)2q^n=q^(n+1)+q^(n+2)

Also divide by q n

2=q+q^2

q^2+q-2=0

q+2)(q-1)=0

q=-2 or q=1 (rounded).

So q=-2

Analysis: An is a proportional series, the common ratio = q, let the first term = a1, then sn=a1(1-q n) (1-q).

s(n+1)=a1[1-q (n+1)] (1-q),s(n+2)=a1[1-q (n+2)] (1-q), and s(n+1), sn,s(n+2) into an equal difference series, 2sn=s(n+1)+s(n+2), that is, 2a1(1-q n) (1-q)=a1[1-q (n+1)] (1-q)+

a1[1-q (n+2)] (1-q),2q n=q (n+1)+q (n+2), q n≠0,2=q+q 2, q=1, rounded, q=2

q is equal to 1, not true; When q is not equal to 1, 2sn=sn+1 + sn+2, substitute the summation formula to get the equation about q, solve, q=-2

From the condition, we can know that 2SN=SN+1+SN+2

Sorted out sn+2-sn=sn-sn+1

We can get -an+1 (n+1 is the subscript) = an+2 (n+2 is the subscript) + an+1 (n+1 is the subscript).

An+2 (n+2 is the subscript) = -2 (an+1) (n+1 is the subscript), so the common ratio q is -2

s(n+1), sn, and s(n+2) are in a series of equal differences.

That is, sn+2 + sn+1=2sn

sn+2 - sn=sn - sn+1

a(n+2)+a(n+1)= -a(n+1)a(n+2)=-2*a(n+1)

So q=-2

Assuming n=1, then there is a series of equal differences between s2, s1, and s3, and there is -a2=a2+a3

So a3 a2=-2

So q=-2

Let the tolerance be d, and the ratio of [bn} is q

then s2=3+3+b=6+b, s3=s2+3+2b=9+3b, b2=q, b3=q 2

and b2*s2=64, b3*s3=960

So q(6+b) = 64

q^2(9+3b)=960

q=8, b=2

The tolerance d and the common ratio q of {an} and {bn} (q=1 and q≠1 in two cases) are set, and then the recursive formula and the first n terms and formulas (as should be in the book) are brought into the solution.

The reciprocal of x, y, and z is a series of equal differences.

1/x=a-d x=1/(a-d)

1/y=a y=1/a

1/z=a+d z=1/(a+d)

x-y/y-z

1 (a-d)-1 a) (1 a-1 (a+d))=(a+d) (a-d) The value is indefinite.

x=1 2 y=1 3 z=1 4 The reciprocal is a series of 2,3,4 equal differences, x-y y-z=2

x=1 3 y=1 4 z=1 5 The reciprocal is a series of 3,4,5 equal differences, x-y y-z=5 4