Question

Prove that for any four consecutive terms of an arith-
metic sequence, the sum of the two terms on the two
ends 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.