Question

prove that for any four consecutive terms of an arthimetic sequance, the sum of two terms on the two end and the sum of the two terms in the middle are the same​

Answer 1

Answer:

Let the smallest term be a and the common difference be d. Then the four consecutive terms are a, a+d, a+2d and a+3d. The two terms on the end sum up to a+(a+3d) = 2a+3d and the two terms in the middle sum up to (a+d)+(a+2d)=2a+3d. Hence they are the same.