It is known that the arithmetic sequence {an} satisfies A1 = 3. A4 + A8 = 26, and the sum of the first n terms of {an} is SN

It is known that the arithmetic sequence {an} satisfies A1 = 3. A4 + A8 = 26, and the sum of the first n terms of {an} is SN

Let the tolerance be d
∵a1=3,a4+a8=26,
∴a4+a8=a1+3d+a1+7d=2a1+10d=26
We can get d = 2,
That is to say, an = a1 + (n-1) d = 2n + 1
Then Sn = A1N + n (n-1) d / 2 = N2 + 2n

If {an} is an arithmetic sequence, Sn is the sum of the first n terms. If A1 > 0, the maximum natural number n of d0 is_________

11

Let the sum of the first n terms of the arithmetic sequence {an} be SN. If A1 > 0 and S4 = S8, then when Sn reaches the maximum, the value of n is ()
A. 5B. 6C. 7D. 8

From S4 = S8: 4A1 + 4 × 32D = 8A1 + 8 × 72d, the solution is: A1 = - 112d, and A1 > 0, get d < 0, so Sn = Na1 + n (n − 1) 2D = d2n2 + (a1-d2) n, from d < 0, get SN is an open downward parabola about n, and S4 = S8, from the symmetry of quadratic function, when n = 4 + 82 = 6, Sn gets the maximum value

In the arithmetic sequence {an}, A1 = - 25, S3 = S8, then the minimum value of the first n terms and Sn is ()
A. -80B. -76C. -75D. -74

From S3 = S8, we can get 3 × (- 25) + 3 × 22d = 8 × (- 25) + 8 × 72d, ∧ d = 5, ∧ Sn = - 25N + n (n − 1) 2 × 5 = 52n2-55n2; the image of Sn is shown in the figure, and its symmetry axis is x = 5.5, ∧ when n = 5 or n = 6, the minimum value of Sn is - 75

Given that A1 = - 25, S3 = S8 in the arithmetic sequence {an}, then when an > 0, the smallest positive integer n=

S3=S8
3a1+3*2d/2=8a1+8*7d/2
3a1+3d=8a1+28d
5a1+25d=0
5(-25)+25d=0
d=5
an=a1+(n-1)d=-25+(n-1)5>0
n-1>5
n>6
n=7

In the arithmetic sequence {an}, A1 = - 25, S3 = S8, then the minimum value of the first n terms and Sn is ()
A. -80B. -76C. -75D. -74

From S3 = S8, we can get 3 × (- 25) + 3 × 22d = 8 × (- 25) + 8 × 72d, ∧ d = 5, ∧ Sn = - 25N + n (n − 1) 2 × 5 = 52n2-55n2; the image of Sn is shown in the figure, and its symmetry axis is x = 5.5, ∧ when n = 5 or n = 6, the minimum value of Sn is - 75

In the arithmetic sequence {an}, A1 = - 25, S3 = S8, then the minimum value of the first n terms and Sn is ()
A. -80B. -76C. -75D. -74

From S3 = S8, we can get 3 × (- 25) + 3 × 22d = 8 × (- 25) + 8 × 72d, ∧ d = 5, ∧ Sn = - 25N + n (n − 1) 2 × 5 = 52n2-55n2; the image of Sn is shown in the figure, and its symmetry axis is x = 5.5, ∧ when n = 5 or n = 6, the minimum value of Sn is - 75

Let Sn be the sum of the first n terms of the equal ratio sequence [an], S3, S9 and S6 be equal difference sequence, and A2 plus A5 is equal to 2am
Finding the value of M

First item a
2S9=2a(q^9-1)/(q-1)
S3+S6=a(a^3-1)/(q-1)+a(a^6-1)/(q-1)
2S9=S3+S6
Obviously, a is not equal to 0
2(q^9-1)=a^3-1+q^6-1
2q^9=q^3+q^6
2q^7=q+q^4
2aq^7=aq+aq^4
2a8=a2+a5
So m = 8

It is known that SN is the sum of the first n terms of the equal ratio sequence {an}, S3, S9 and S6 are equal difference sequence, and A2 + A5 = 2A, then M=
A2 + A5 = 2am missed ·~

S3=a1(1-q^3)/（1-q),S9=a1(1-q^9)/(1-q).S6=a1(1-q^6)/(1-q）
If it is an arithmetic sequence, then q ^ 3 + Q ^ 6 = 2 * q ^ 9. If it is divided by Q ^ 2, then q + Q ^ 4 = 2 * q ^ 7

If the sum of the first n terms of the arithmetic sequence {an} is Sn, A5 = 11, S12 = 186, then A8 = ()
A. 18B. 20C. 21D. 22

From the properties of sequence, we get that a1 + A12 = A5 + A8, because S12 = & nbsp; 122 × (a1 + A12) = 186, so a1 + A12 = A5 + A8 = 31, because A5 = 11, so A8 = 20, so we choose B