What is the sum of the first $42$ terms of the sequence?

Question

The sum of the first $14$ terms of an arithmetic sequence is $2014$. If the sum of the first $28$ terms of the same sequence is $2014$, what is the sum of the first $42$ terms of the sequence?

I thought the sum would be the same ($2014$) since it didn't change from the $14$ term sum to the $28$ term sum.

Answer 1

The sum of the first 14 terms is $2014$, the sum of the next 14 terms is $0$, and then the sum of the next 14 terms is $-2014$.

And the sum of $$2014 + 0 + (-2014) = 0$$

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This is because the sequence of $$\begin{array}{ccccc} (a+0d)& +&(a+1d)& + \cdots +& (a+13d),\\ (a+14d)& +&(a+15d)& + \cdots +& (a+27d),\\ (a+28d)& +&(a+29d)& + \cdots +& (a+41d),\\ &&\vdots \end{array}$$ are in AP with term difference $14^2 d$.

Answer 2

Hint: If you group terms of an arithmetic sequence $k$ at a time (in non-overlapping groups), and you form a sequence of the sum of those groupings of $k$ terms, that new sequence will be arithmetic.

Answer 3

Hint: since the sum of terms $15$ to $28$ is zero, so is their average, which is the value of term $"21\frac 12"=\cfrac {15+28}2$. Let the difference between successive terms be $2d$ - you should be able to write down terms $21$ and $22$ and work from there. Having found that zero value, exploit the symmetry.