The first n terms and the formula s (n) = n * a (1)+n * (n-1) * d/2 or S(n)=n*(a( 1)+a(n))/2 n are positive integers. 2. From the definition and general formula of arithmetic progression, we can also deduce the first n terms and formulas: a (1)+a (n) = a (2)+a (n-1) = a (3)+a (n-2) = … = a (k)+. (similar: p (1)+p (n) = p (2)+p (n-1) = p (3)+p (n-2) =) ... = p (k)+p (n-k+65433) If m+n=2p, then a (m)+a (n) = 2 * a (p) (proof of 3: p (m)+p (n) = b (0)+b (1) * m+b (0)+b (65438). Because m+n=p+q, p(m)+p(n)=p(p)+p (q)) IV. Other inferences ① sum = (first item+last item) × number of items ÷2 (proof: s (n) = [n, n 2] *. 0， 1/2]*[b(0)； b( 1)]=n*b0+ 1/2*b 1*n+ 1/2*b 1*n^2(p( 1)+p(n))* n/2 =(b(0)+b( 1)+b(0)+b( 1)* n * n/2 = n * B0+1* n+1/2 * b1* b1* n 2 = s (n)) Number of items = (last item-first item) ÷ tolerance ④ Last item = first item+(item) Then there is a(m)+a(n)=a(p) +a(q), such as a (m)+a (n) = a (1)+(m-1) * d+a (1)+. A (1) and d are constants, so if m, n, p, q∈N* and m+n=p+q, there is a(m)+a(n)=a(p)+a(q) Note: 1 When the number of terms is (2n- 1) (n), S-S = a, =. (3) If the sequence is arithmetic progression, then S n, S2n -Sn, S3n-S2n, … still become arithmetic progression with a tolerance of k 2d. (4) If the sum of the first n terms of two arithmetic progression is S and T respectively (n is an odd number). S = b (n > m), then S = (a-b). [6] In arithmetic progression, it is a linear function of n, and all points (n,) are on the straight line Y = x+(a-). [7] Remember that the sum of the first n terms of arithmetic progression is S1. If a > 0, the tolerance D 0, s is the minimum when a ≤0 and a ≥0.