The sum of the first n terms of the sequence {an} is SN. It is known that S1, S3 and S2 are isometric sequences (I) find the common ratio Q of {an}; (II) A1-A3 = 3, find SN

The sum of the first n terms of the sequence {an} is SN. It is known that S1, S3 and S2 are isometric sequences (I) find the common ratio Q of {an}; (II) A1-A3 = 3, find SN

(I) ∵ the sum of the first n terms of the sequence {an} is SN,
S1, S3 and S2 are in arithmetic sequence,
∴2（a1+a1q+a1q2）=a1+a1+a1q，
The solution is q = - 1
2 or q = 0
∴q=-1
2．
（Ⅱ）∵a1-a3=3，q=-1
2，
∴a1−1
4a1＝3，a1=4，
∴Sn＝4[1−(−1
2)n]
1+1
2=8
3[1-（-1
2）n]．

Let S3 = 12, and 2A1, A2, A3 + 1 be equal proportion sequence, find SN

Let the tolerance of arithmetic sequence {an} be d
a22＝2a1(a3+1)
3a1+3×2
2D = 12
a1＝1
D = 3 or
a1＝8
d＝−4 ，
∴sn=1
2n (3n-1) or Sn = 2n (5-N)

If the sum of the first n terms is Sn in the equal ratio sequence {an}, if A2, A4 and A3 are equal difference sequence, we can judge whether S2, S4, S3 are equal difference sequence, and give the proof

If A2, A4, A3 is the arithmetic sequence, then 2a4 = A2 + a3, therefore, 2A2 * q ^ 2 = A2 + A2 * q is 2q ^ 2-q-1 = 0, so q = - 1 / 2 or q = 1 (1) if q = -1 / 2, then S2 = a1 + A2 = a1-a1 / 2 = A1 / 2s3 = S2 + a3 = A1 / 2 + A1 * (- 1 / 2) ^ 2 = 3A1 / 4s4 = S3 + A4 = 3A1 / 4 + A1 * (- 1 / 2) ^ 3 = 3A1 / 4s4 = S3 + A4 = 3A1 / 4 + A1 * (- 1 / 2) ^ 3 = 5A1 / 8, therefore, therefore, therefore, therefore, therefore, therefore, we can make a 4 = 3A1 / 4 = 1 / 2 + 1 / 2) ^ 3 = 5A1 / 8, so 2s4 = S2 + S3 (2) if q = 1, S2 =

The sum of the first n terms of the sequence {an} is SN. It is known that S1, S3 and S2 are isometric sequences. (1) find the common ratio Q of {an} (2) if A1-A3 = 3, find SN

2S3=S1+S2
Then 2 (a1 + A2 + a3) = a1 + (a1 + A2)
2a1+2a1q+2a1q²=2a1+a1q
2q²+q=0
Obviously, Q ≠ 0
So q = - 1 / 2
a1-a3=a1-a1/4=3
a1=4
So Sn = 4 [1 - (- 1 / 2) ^ n] / (1 + 1 / 2)
=8[1-(-1/2)^n]/3

The sum of the first n terms of the sequence {an} is SN. It is known that S1, S3 and S2 are isometric sequences (I) find the common ratio Q of {an}; (II) A1-A3 = 3, find SN

(I) ∵ the sum of the first n terms of the sequence {an} is SN,
S1, S3 and S2 are in arithmetic sequence,
∴2（a1+a1q+a1q2）=a1+a1+a1q，
The solution is q = - 1
2 or q = 0
∴q=-1
2．
（Ⅱ）∵a1-a3=3，q=-1
2，
∴a1−1
4a1＝3，a1=4，
∴Sn＝4[1−(−1
2)n]
1+1
2=8
3[1-（-1
2）n]．

It is known that the sequence {an} is an arithmetic sequence with tolerance not 0, and A1, A3, A4 are proportional sequence, and Sn is the sum of the first n terms of sequence {an}, and the value of S3 / S5 is calculated

A1, A3 and A4 are proportional sequence,
∴（a1+2d)^2=a1(a1+3d),
a1d+4d^2=0,d≠0,
∴a1=-4d.
∴S3=-12d+3d=-9d,
S5=-20d+10d=-10d,
∴S3/S5=9/10.

It is known that the sum of the first n terms of the arithmetic sequence {an} is Sn, the tolerance D ≠ 0, and S3 = 9, A1, A3, a7 form an equal proportion sequence (1) Find the general term formula of sequence {an}; (2) Let BN = 2 an, find the first n terms and TN of the sequence {BN}

(1) ∵ A1, A3, a7 are proportional sequence
∴a32=a1a7，
That is (a1 + 2D) 2 = A1 (a1 + 6D),
D = 1
2A1, d = 0 (omitted)
∴S3=3a1+3×2
2×1
2a1=9
2a1=9, a1=2, d=1
ν an = a1 + (n-1) d = 2 + (n-1) = n + 1, that is an = n + 1
（2）∵bn=2an=2n+1，∴b1=4，bn+1
bn＝2．
｛ BN} is an equal ratio sequence with 4 as the first term and 2 as the common ratio,
∴Tn=4(1−2n)
1−2=2n+2-4．

(2011. Jinzhou three mode) the arithmetical sequence {an} with a tolerance of 0 is not equal to 0, and A1 is equal to A3 and A4. The value of S5 − S3 is () A. 2 B. 3 C. 1 Five D. It doesn't exist

Because {an} is an isochromatic sequence, which is proportional to A1, A3 and A4, A32 = a1a4, i.e. (a1 + 2D) 2 = A1 (a1 + 3D),
D (a1 + 4D) = 0 is obtained by simplification, and a1 + 4D = 0 is obtained from D ≠ 0, so A1 = - 4D is A5 = 0,
Then S3 − S2
S5−S3=a3
a4+a5=a1+2d
a1+3d+0=−2d
−d=2
Therefore, a

Let {an} be the number of arithmetic sequence with tolerance not 0, A1 = 2, and A1, A3, A6 are proportional sequence, then the first n terms of {an} and Sn = ()

The tolerance is d
a3=2+2d
a6=2+5d
If the number is equal, then
a3^2=a1*a6
(2+2d)^2=2(2+5d)
4d^2+8d+4=4+10d
4d^2-2d=0
2d(2d-1)=0
D = 1 / 2 (because D is not 0)
an=a1+(n-1)d=2+(n-1)*1/2=3/2+n/2
Sn=(a1+an)n/2=(2+3/2+n/2)n/2=7n/4+n^2/4

It is known that SN is the sum of the first n terms of the arithmetic sequence {an} with tolerance not 0, and S1, S2, S3 are proportional sequence, then A2 + a3 of A1 is equal to the solution step

If the data is wrong, let's give you an idea. Let A1 be the first term of the arithmetic sequence, and the tolerance is d (2A1 + D) 2 = A1 * (3A1 + 3D) 4A1 2 + 4a1d + D? 2 = 3A1? 2 + 3a1da1? 2 + a1d + D? 2 = 0. We can get the quadratic equation about D / A1 by dividing A1? Together. There is no solution to the quadratic equation of this problem = (A2 + a3) /