In the arithmetic sequence, A1 = 6 and S4 = S9 are known, then when n is the value, what is the maximum value of Sn?

In the arithmetic sequence, A1 = 6 and S4 = S9 are known, then when n is the value, what is the maximum value of Sn?

S4 = 4A1 + 4 * 3 * D / 2 = 4A1 + 6ds9 = 9a1 + 9 * 8 * D / 2 = 9a1 + 36ds4 = S9 -- > 4A1 + 6D = 9a1 + 36d5a1 = - 30da1 = - 6dsn = Na1 + n (n-1) d / 2 = - 6nd + n (n-1) D / 2 = D / 2 (n ^ 2-n-12n) = D / 2 ((N-13 / 2) ^ 2-169 / 4). Therefore, when n = 6 or n = 7, Sn takes the maximum value

In the arithmetic sequence {an}, A1

Analysis:
S12=S9+(a10+a11+a12)=S9
∴a10+a11+a12=0
This is an arithmetic sequence,
∴a10+a12=2a11
∴a11=0
∵a10
From A1 to A10 are less than 0, and from A12 are greater than 0
The minimum value of Sn is S10 or S11 (because S11 = S10 + a11 = S10)
Sn has no maximum,

In the arithmetic sequence an, A1

s9=s12-->a10+a11+a12=0-->a11=0;
So S10 and S11 are the smallest
（an

In the arithmetic sequence {an}, A1 ＞ 0, S4 = S9, then when Sn takes the maximum value, n=______ ．

4A1 + 4 × 32D = 9a1 + 9 × 82d, the solution is A1 = - 6D.  Sn = − 6dn + n (n + 1) D2 = d2n2 − 13d2n = D2 (n − 132) 2 − 1698d,  A1 > 0, d < 0,  when n = 6 or 7, the maximum value of Sn is - 1698d

In the arithmetic sequence {an}, A1 ＞ 0, S4 = S9, then when Sn takes the maximum value, n=______ ．

4A1 + 4 × 32D = 9a1 + 9 × 82d, the solution is A1 = - 6D.  Sn = − 6dn + n (n + 1) D2 = d2n2 − 13d2n = D2 (n − 132) 2 − 1698d,  A1 > 0, d < 0,  when n = 6 or 7, the maximum value of Sn is - 1698d

In the arithmetic sequence an, A1 = 25, Sn = S9, find the maximum value of Sn

Sn = [(a1 + an) n] / 2, that is 118 = n (25 + an)
And 118 = 2 × 59 n is a positive integer
So n = 2 or n = 59
1) When n = 2, the sequence has only two terms, A1 = 25, A2 = 34
The maximum SN is S2 = 59
2) When n = 59, an = - 23 d = - 24 / 29 has 59 items
An decreasing
a31=5/29 a32=-19/29
So the maximum SN is S31 = 11315 / 29

In the arithmetic sequence {an}, it is known that A1 > 0, Sn is the sum of the first n terms of the sequence, if S9 > 0, S10

It's not as troublesome as quadratic function
s9=9a1+(9*8/2)d=9a1+36d>0,a1>-4d,
s10=10a1+45d

The sum of the preceding terms of the arithmetic sequence {an} is SN. If S90 is known, the minimum value of the sum of the preceding n terms of the sequence is n

a(n)=a+(n-1)d,
d> At 0, a (n) increases monotonically,
ds(9)=9[a+4d],0a+4d,0< a+9d/2,
d/2 > a+4d+d/2 = a+9d/2 > 0.
d> A (n) increases monotonically
-9d/2 < a < -4d < 0.
a(5)=a+4da+9d/2>0.
a(1)0,s(6)>s(5).
So there are always,
When n is not 5, s (n) > s (5)
When n = 5, s (n) is the minimum

SN is used to represent the sum of the first n terms of the arithmetic sequence an, if S9 = 0, S10 = - 5, then A1 =?,

Sn = Na1 + n (n-1) d / 2Sn / N = a1 + (n-1) d / 2 = D / 2 * n + (a1-d / 2) so S9 / 9 = D / 2 * 9 + (a1-d / 2) = 0 (1) S10 / 10 = D / 2 * 10 + (a1-d / 2) = - 1 / 2 (2) (2) - (1): D / 2 = - 1 / 2 D = - 1 generation (1): 8 * D / 2 + A1 = 0, A1 = - 8 * (- 1 / 2) = 4, so D = - 1, A1 = 4

Let the sum of the first n terms of the arithmetic sequence {an} be SN. If S9 ＞ 0 and S10 ＜ 0, then 2A1, 22a2 The largest of 29a9 is ()
A. 2a1B. 25a5C. 26a6D. 29a9

∵ S9 = 9 (a1 + A9) 2 = 9A5 > 0, S10 = 10 (a1 + A10) 2 = 5 (a1 + A10) ∵ A5 > 0, A5 + A6 < 0, A6 < 0 ∵ in the arithmetic sequence {an}, A1 > A2 > A3 > A4 > A5 > 0 > A6