How do you do this problem? Find the rules and draw five

How do you do this problem? Find the rules and draw five

The following five are: 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000

How to draw on the problem of finding rules

In the arithmetic sequence {an}, A1 ＞ 0, D ≠ 0, if S3 = S11, then the maximum value of Sn is ()

S3=3a1+3*2/2d=3a1+3d
S11=11a1+11*10/2d=11a1+55d
When S3 = S11, 3A1 + 3D = 11A1 + 55D, A1 = - 13 / 2D
Sn=na1+n(n-1)d/2=n(n-1)d/2-13nd/2=d/2*(n^2-14n)
Because A1 > 0, D

Let Sn be the sum of the first n terms of the arithmetic sequence {an}. If S3 / S6 = 1 / 3, then S6 / S12 is equal to——

SN is the sum of the first n terms of the arithmetic sequence {an}
Then S3, s6-s3, s9-s6, s12-s9 also form arithmetic sequence
Because S3 / S6 = 1 / 3
So S6 = 3s3
So s6-s3 = 2s3
So s9-s6 = 3s3, s12-s9 = 4S3
So S9 = S6 + 3s3 = 6s3
S12=S9+4S3=10S3
So S6 / S12 = 3s3 / 10s3 = 3 / 10
If you don't understand, please hi me, I wish you a happy study!

Let Sn be the sum of the first n terms of the arithmetic sequence {an}, if s3s6 = 13, then s6s12 = ()
A. 310B. 13C. 18D. 19

Let the first term of the arithmetic sequence {an} be A1 and the tolerance be d. from the summation formula of the arithmetic sequence, we can get s3s6 = 3A1 + 3d6a1 + 15d = 13, A1 = 2D and D ≠ 0, s6s12 = 6A1 + 15d12a1 + 66d = 27d90d = 310

SN is the sum of the first n terms of the arithmetic sequence an, if S3 / S6 = 1 / 3, then S6 / S12 =?

SN is the sum of the first n terms of an arithmetic sequence, so S3, s6-s3, s9-s6, s12-s9 are also arithmetic sequences. Because S3 / S6 = 1 / 3, S6 = 3s3, so s6-s3 = 2s3, so the tolerance is d = (s6-s3) - S3 = S3, so s9-s6 = S3 + 2s3 = 3s3s3, s12-s9 = S3 + 3s3 = 4S3, so S9 = S6 + 3s3 = 6s3s12 = S9 + 4S3 = 10s3, so S6 / S12 = 3s3 / 1

The sum of the first n terms of the arithmetic sequence {an} is SN_ The square of 1 + am + 1-am equals zero. S2m-1 equals 38. Then M equals 0

Am-1+Am+1-Am^2=0
Because am is an arithmetic sequence, am-1 + am + 1 = 2am
So 2am am ^ 2 = 0, so am = 2 or 0
S2m-1 = (2m-1) am = 38 {note: am is the median of 2m-1}
So am = 0, so am = 2
So m = 10

If M > 1, am-1 + am + 1-am2 = 0, s2m-1 = 38, then M is equal to ()
A. 38B. 20C. 10D. 9

According to the property of arithmetic sequence, we can get: am-1 + am + 1 = 2am, then am-1 + am + 1-am2 = am (2-AM) = 0, the solution is: am = 0 or am = 2, if am is equal to 0, it is obvious that s2m-1 = (2m − 1) (a1 + A2M − 1) 2 = (2m-1) am = 38 does not hold, so am = 2, s2m-1 = (2m-1) am = 4M-2 = 38, the solution is m = 10

Given the first n terms of the sequence {an} and Sn = n + 1n + 2, then A3 = ()
A. 120B. 124C. 128D. 132

A3 = s3-s2 = 3 + 13 + 2-2 + 12 + 2 = 120

Given that the sum of the first n terms of a sequence an is Sn = 2An + 1, then A3 is equal to

Because Sn = 2An + 1, S1 = 2A1 + 1, S1 = A1, that is, A1 = 2A1 + 1, A1 = - 1
Sn=2an+1 (1)
S(n-1)=2a(n-1)+1 (2)
(1) - (2) get
an=2an-2a(n-1)
an=2a(n-1)
So an is an equal ratio sequence with prime minister - 1 and common ratio 2
a3=a1*2^2=-1*4=-4