Can these numbers give a mathematical formula? Arithmetic sequence

Can these numbers give a mathematical formula? Arithmetic sequence

Each of these numbers is 6 larger than the previous one, so the tolerance D is 6
So the first term A1 = 0, the nth term an = a1 + (n-1) d = 0 + (n-1) × 6 = 6n-6
Sum of the first n terms Sn = n (a1 + an) / 2
But your eighth item is not an arithmetic sequence

Arithmetic sequence 30,24,18,12 What is the general term formula of

d=-6
a1=30
an=a1+(n-1)*d
an=30+(n-1)*(-6)
an=-6*n+36

Please answer the arithmetic sequence in senior one, thank you! (18 21:44:45)
In the arithmetic sequence with 2n + 1 terms, if the sum of all odd terms is 165 and the sum of all even terms is 150, then n is equal to (& # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160;)
The known sequence {an} satisfies A0 = 1, an = A0 + A1 + +An-1 (n ≥ 1), then when n ≥ 1, an = (& # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160;)
The tolerance D in the arithmetic sequence {an} is not equal to 0. If A1, A3 and A9 are proportional sequence, then the value of (a1 + a3 + A9) / (A2 + A4 + A10) is equal to (& # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160; & # 160;)

The even term is one less than the odd term, so there should be a1 + DN = 165-150 = 15, that is, a [n + 1] = 15;
S [2n + 1] = (2n + 1) * a [n + 1] = 15 * (2n + 1) = 165 + 150 = 315, so n = (315 / 15-1) / 2 = 10;
Note Sn = A0 + A1 + +So Sn = 2 * s [n-1], and S1 = A0 + A1 = 1 + 1 = 2;
So Sn = 2 ^ n; an = s [n-1] = 2 ^ (n-1);
A1 * A9 = A3 * A3, that is, A1 (a1 + 8D) = (a1 + 2D) ^ 2 and D ≠ 0, so d = A1, so an = n * A1;
So (a1 + a3 + A9) / (A2 + A4 + A10) = (1 + 3 + 9) A1 / (2 + 4 + 10) A1 = 13 / 16