If the sum of the first n terms of the arithmetic sequence {an} is Sn and 6s5-5s3 = 5, then A4=______ ．

If the sum of the first n terms of the arithmetic sequence {an} is Sn and 6s5-5s3 = 5, then A4=______ ．

∵ Sn = Na1 + 12n (n-1) d ∵ S5 = 5A1 + 10d, S3 = 3A1 + 3D ∵ 6s5-5s3 = 30a1 + 60d - (15a1 + 15d) = 15a1 + 45d = 15 (a1 + 3D) = 15a4 = 5, the solution is A4 = 13, so the answer is: 13

If the sum of the first n terms of the arithmetic sequence {an} is Sn and 6s5-5s3 = 5, then A4=______ ．

∵ Sn = Na1 + 12n (n-1) d ∵ S5 = 5A1 + 10d, S3 = 3A1 + 3D ∵ 6s5-5s3 = 30a1 + 60d - (15a1 + 15d) = 15a1 + 45d = 15 (a1 + 3D) = 15a4 = 5, the solution is A4 = 13, so the answer is: 13

If the sum of the first n terms of the arithmetic sequence {an} is Sn and 6s5-5s3 = 5, then A4=______ ．

∵ Sn = Na1 + 12n (n-1) d ∵ S5 = 5A1 + 10d, S3 = 3A1 + 3D ∵ 6s5-5s3 = 30a1 + 60d - (15a1 + 15d) = 15a1 + 45d = 15 (a1 + 3D) = 15a4 = 5, the solution is A4 = 13, so the answer is: 13

The sum of the first n terms of the arithmetic sequence an is Sn, and 6sn-5s3 = 5, A4=
Wrong, 6s5-5s3 = 5, A4

The title is wrong
6S5-5S3=5
6(5a1+10d)-5(3a1+3d)=5
15a1+45d=5
a1+3d=1/3
That is, A4 = 1 / 3

If a1 + A2 = 5, A3 + A4 = 9, then the value of S10 is () a. 55B. 60C. 65D. 70

In ∵ arithmetic sequence {an}, a1 + A2 = 5, A3 + A4 = 9, ∵ a1 + A1 + D = 5A1 + 2D + A1 + 3D = 9, the solution is A1 = 2, d = 1, ∵ S10 = 10 × 2 + 10 × 92 × 1 = 65

In the arithmetic sequence {an}, if a1 + A4 = 10, A2-A3 = 2, then the first n terms and Sn of the sequence are ()
A. n2+7nB. 9n-n2C. 3n-n2D. 15n-n2

Let the first term of the arithmetic sequence {an} be A1, the tolerance be D, ∵ a1 + A4 = 10, A2-A3 = 2, ∵ d = - 2, A1 = 8, ∵ the first n terms of the sequence and Sn = Na1 + n (n − 1) 2D = 9n-n2, so choose B

The sum of the first n terms of the arithmetic sequence an is Sn, and A2 equals negative 5, A4 minus A6 plus 6 equals 0;
(3) When SN is greater than 0, the minimum value of n is obtained

(1) A4-a6 + 6 = a1 + 3D - (a1 + 5d) = 6 = - 2D + 6 = 0, so d = 3; and A2 = a1 + D = - 5, so A1 = - 8 (2) S10 = 10 * 2 + 10 * (10-1) * - 8 / 2 = - 340 (3) Sn = - 8N + n (n-1) * 3 / 2 = n & # 178; 3 / 2 + 19n / 2 ＞ 0, the solution is n ＞ 19 / 3 or n ＜ 0 (rounding off)

It is known that Sn in the arithmetic sequence {an} is the sum of its first n terms, let A6 = 2, S10 = 10

Sn=n(a1+an)/2
an=a1+(n+1)d
According to these two formulas, you can find what you want

It is known that the sum of the first n terms of the arithmetic sequence an is SN. If A6 = 18-a5, then S10 is equal to SN

90
S10=A1+A2+A3+A4+ A10
A1+A10=A2+A9=A3+A8=A4+A7=A5+A6
Because A5 + A6 = 18
Solid S10 = 90

Let Sn be the sum of the first n terms of the arithmetic sequence {an}, if s3s6 = 13, then s6s7=______ ．

Let the first term of the arithmetic sequence be a, and the tolerance be D, ∵ S3 = 3A + 3ds6 = 6A + 15ds7 = 7a + 21d ∵ s3s6 = 13, ∵ 3A + 3d6a + 15d = 13 ∵ a = 2D ∵ s6s7 = 6A + 15d7a + 21d = 2735, so the answer is: 2735