The sum of the 1st and 2nd terms of an arithmetic series is $x$ and the sum of the $(n-1)$th and $n$th terms is $y$. Show that the sum of the first $n$ terms is $(n/4)(x+y)$. Find an expression for the common difference in terms of $x$, $y$ and $n$.
I have tried substituting the terms into the expression for the sum of an arithmetic series, then manipulating them and equating them in different ways but I have not got close to an answer for either part of the question.
The answer in the book for the common difference is $\dfrac{y-x}{2(n – 2)}$.
Best Answer
Hint: Let $d$ be the common difference. Then observe that: $$ x = t_1 + t_2 = t_1 + (t_1 + d) = 2t_1 + d \\ y = t_{n-1} + t_n = (t_n - d) + t_n = 2t_n - d $$ For the first part, consider adding the two equations then dividing by two. For the second part, consider subtracting the two equations then dividing by two.