In the arithmetic sequence, A2 + A5 = 11, A3 + A6 = 17, find the general formula of {an}

In the arithmetic sequence, A2 + A5 = 11, A3 + A6 = 17, find the general formula of {an}

(a3+a6)-(a2+a5)=6
a3-a2+a6-a5=6
2k=6
So the tolerance k = 3
A2 + A5 = a1 + K + A1 + 4K = 2A1 + 5K = 2A1 + 15 = 11, so Prime Minister A1 = - 2
So the general formula of sequence {an} is an = - 2 + 3 (n-1) = 3n-5

[mathematics of the first year of senior high school] it is known that f (x + 1) = x quartic power + 4, in the arithmetic sequence {an}, A1 = f (x-1), A2 = - 3 / 2, A3 = f (x). (1). Find x (2). Find the value of A2 + A5 + A8 +... + A26

X has no solution let x + 1 = t, then x = T-1 f (x + 1) = x ^ 4 + 4 f (T) = (t-1) ^ 4 + 4  f (x) = (x-1) ^ 4 + 4 > 0. According to the {an} a1 + a3 = 2A2 (X-2) ^ 4 + 4 + (x-1) ^ 4 + 4 = 2 * - 3 / 2 = - 3, this equation does not hold in R. this problem has no solution

The arithmetic sequence {an} has 2n + 1 terms, where a1 + a3 + +a2n+1=4，a2+a4+… +A2N = 3, then the value of n is ()
A. 3B. 5C. 7D. 9

The arithmetic sequence {an} has 2n + 1 terms, ∵ a1 + a3 + +a2n+1=4，a2+a4+… +A2N = 3, subtracting the two formulas, we can get a1 + nd = 1, adding the two formulas, we can get s2n + 1 = 7 = (2n + 1) a1 + (2n + 1) · 2n2d, ■ (2n + 1) (a1 + nd) = 7 ■ (2n + 1) = 7, ■ n = 3. So we choose a