The sum of the first four items of the arithmetic sequence is - 8, and the sum of the first five items is 85

The sum of the first four items of the arithmetic sequence is - 8, and the sum of the first five items is 85

It can be found that A5 = 93
Let A1 = 93-4d, A2 = 93-3d, A3 = 93-2d, A4 = 93-d
The sum of the first four terms = 372-10d = - 8, and the tolerance d = 38
So the first term, A1 = 93-4 × 38 = - 59

It is known that the sum of the first term and the fourth term of the arithmetic sequence is 10, and the difference between the second term and the third term is 2. Find the sum of the first n terms of the arithmetic sequence
Here's an additional question. Friends who can do it will answer it together!
Given that the recurrence formula of a sequence is a (subscript n + 1) = 3A n, A1 = - 1 / 3, then the general term formula of the sequence is?

1. Equal difference: the general formula an = a1 + (n-1) DA1 = A1, A2 = a1 + da3 = a1 + 2D, A4 = a1 + 3D is known as a1 + A4 = 10, that is, 2A1 + 3D = 10 is known as A2-A3 = 2, that is, a1 + D - (a1 + 2D) = 2, so d = - 2, A1 = 8, an = a1 + (n-1) d = 8-2 (n-1) sum: SN = n (a1 + an) / 2 = Na1 + n (n-1) d / 2 = 8n-n ^ 2 + n = 9N -

Arithmetic sequence - 3, 1, 5 Item 15 of is ()
A. 40B. 53C. 63D. 76

From the question: A1 = - 3, d = 1 - (- 3) = 4. An = a1 + (n-1) d = 4n-7. Substituting n = 15, A15 = 4 × 15-7 = 53