In the arithmetic sequence {an}, 3 (A3 + A5) + 2 (A7 + A10 + A13) = 24, then the sum of the first 13 items in the sequence is () A. 13B. 26C. 52D. 56

In the arithmetic sequence {an}, 3 (A3 + A5) + 2 (A7 + A10 + A13) = 24, then the sum of the first 13 items in the sequence is () A. 13B. 26C. 52D. 56

From the properties of the arithmetic sequence, we can get: A3 + A5 = 2a4, a7 + A13 = 2a10, substituting into the known, we can get 3 × 2a4 + 2 × 3a10 = 24, that is, A4 + A10 = 4, so the sum of the first 13 items of the sequence S13 = 13 (a1 + A13) 2 = 13 (A4 + A10) 2 = 13 × 42 = 26, so we choose B

If {an} is an arithmetic sequence, A3 + a7-a10 = 8, a11-a4 = 4, then the sum of the first 13 terms of the sequence is:
Man, please be more detailed

Because an is an arithmetic sequence, let an = a1 + (n-1) d
So A3 = a1 + 2D A7 = a1 + 6D A10 = a1 + 9D a11 = a1 + 10d A4 = a1 + 3D
Substituting A3 + a7-a10 = 8 and a11-a4 = 4
The solution is A1 = 60 / 7 and d = 4 / 7
So A13 = 108 / 7
And Sn = n (a1 + an) / 2
So S13 = 156

(2011 Zhengzhou three module) number {an}, a3=2, a7=1, if {1an+1} is arithmetic progression, then a11= ().
A. 0B. 12C. 23D. 2

Let the tolerance of the sequence {1An + 1} be d ∵ in the sequence {an}, A3 = 2, a7 = 1, and the sequence {1An + 1} is the arithmetic sequence ∵ 1A7 + 1 = 1A3 + 1 + 4D. Substitute A3 = 2, a7 = 1 into d = 124 ∵ 1a11 + 1 = 1A7 + 1 + 4D ∵ a11 = 12, so B is selected