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.

You can deduce that the sum of terms 15 to 28 is 0. – peterwhy Dec 11 '15 at 20:07 — peterwhy, Dec 11 '15 at 20:07
Right. Then how do I get the sum of the next terms? – Puzzled417 Dec 11 '15 at 20:07 — Puzzled417, Dec 11 '15 at 20:07

peterwhy · Accepted Answer · 2015-12-11T20:31:57.307

3

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 $$2014 + 0 + (-2014) = 0$$

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 $14^2 d$.

edited Dec 11 '15 at 20:31

answered Dec 11 '15 at 20:13

peterwhy

22,256

How did you get term difference $14^2d$? – Puzzled417 Dec 11 '15 at 20:15
1

If you notice there are $14$ values for each row and a constant difference of $14d$, which makes for a total of $14*14d$ which is $(14)^2d$ – JB King Dec 11 '15 at 20:18
Okay, I see that the third row is $14^2*d$ greater than the second row. How does this mean it's sum is $-2014$? – Puzzled417 Dec 11 '15 at 20:21
So the first row adds to $2014$; the second row adds to $0$, which is $2014$ less than the first. And the third row should also adds to be $2014$ less than $0$. – peterwhy Dec 11 '15 at 20:24
Oh I see why now. It's because $14^2d$ is -2014 right? – Puzzled417 Dec 11 '15 at 20:25
@Puzzled417 Right.${}{}{}{}{}{}$ – peterwhy Dec 11 '15 at 20:26
Yes that was a typo. – Puzzled417 Dec 11 '15 at 20:27

score 1 · Answer 2 · answered Dec 11 '15 at 20:12

1

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.

answered Dec 11 '15 at 20:12

paw88789

40,402

That's the easiest way. Also works with overlapping groups. – Dec 11 '15 at 20:39

score 0 · Answer 3 · answered Dec 11 '15 at 20:12

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.

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

3 Answers3