Find the removed term from an arithmetic series, given the sum of the other terms

Question

We are given a sequence made up of pairs of numbers, $(1, 2), (3, 4), …,(x, x+1)$, out of which we must remove one pair. The sum of the numbers which have not been removed is 156. Calculate the sum of the removed numbers.

This is what I got so far:

$(1+2)+(3+4)+(5+6)+...=3+7+11+...$ forms an arithmetic series. Let $S_n$ be the sum of the terms in this series up until the $n^{th}$ term. Let $n$ be the number of pairs.

$S_n=\dfrac{n}{2}(2(3)+(n-1)(4))=n(2n+1)=156+y$, where $y$ is the sum of the removed pair.

Let $y$ be the $a^{th}$ term on the arithmetic series.

$y=3+(a-1)(4)=4a-1$

$\therefore n(2n+1)=156+4a-1 \Rightarrow n(2n+1)=155+4a \Rightarrow a=\dfrac{n(2n+1)-155}{4}$

Since the sum of the numbers is even after $y$ is removed, $n$ must be odd. $a$ must be a positive integer, so $n(2n+1)-155$ must be divisible by 4. Now I'm stuck. The smallest $a$ I found was $4$, which gives $y=15$ and is the correct answer, but I can't understand why it must be the smallest $a$.

Or maybe there is another way of solving this question?

Answer 1

I would rather take the sum without pairing up the summands. Let $n$ be the last number in the sum, i.e., $n=x+1$ (I am doing this so that $n$ is even and so that the number of pairs is $\frac n2$ which is cleaner than working with $x$ in which case the number of pairs is $\frac{x+1}{2}$.

The original sum was

$$1+2+3+4+5+6+\cdots+n-1+n = \frac{n(n+1)}{2}$$ (again, this is cleaner to me than if we had $\frac{(x+1)(x+2)}{2}$).

Now, removing one pair of numbers means we remove some pair $(m, m+1)$, which means we remove $2m+1$, or in other words, we remove an odd number that is not one. That means we need to find some odd number $2m+1$ such that $156+(2m+1)$ will be a triangular number.

Note that this means we need an odd triangular number. The number $\frac{n(n+1)}{2}$ will be odd if neither $n$ nor $n+1$ are divisible by $4$, which happens if $n=4k+1$ or $n=4k+2$ (which, incidentally, means triangular number follow a pattern of two odds, followed by two evens, and then repeat).

From here on, I think some trial and error is the easiest way to go. I tried $n=20$ first because that triangular number is, by feel, close to $200$, and the result looks like will be near there. Of course, $n=20$ won't work because it is not $4k+1$. Neither will $n=19$ for the same reason, and for $n=17$, we have $\frac{n}{n+1} = 153$ which is too low. So $n=18$ is the first reasonable guess, with $n=21$ being the next possibility. Well, luckily, $n=18$ brings us to $\frac{n(n+1)}{2} = 9\cdot 19 = 171 = 156+15=156+(7+8)$, so that's already the solution.

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Note that we can guarantee that we always only have a finitely many solutions. This is because the sum of the numbers that have not been removed is equal to $\frac{n(n+1)}{2} - (2m+1) > \frac{n(n+1)}{2}-2n-1$ which, for large values of $n$, will always be greater than $156$.

In fact, you have $$\frac{n(n+1)}{2}-2n-1 < 156 \iff n\in(\frac12(3-\sqrt{1265}), \frac12(3+\sqrt{1265})\approx(-16.28, 19.28).$$ This tells us that $19$ is the highest possible candidate for $n$.

Answer 2

Since your series consists of positive, increasing terms, there will be at most one possible solution, namely when the series first goes over $156$.

Let $k$ be the smallest value for which $S_k > 156$, and let $S_k - 156 = R>0$. Suppose that $R$ is not a term in the series, i.e. that we cannot get $156$ by removing a term from $S_k$. Consider the next sum $S_{k+1} = S_k + t_{k+1}$. To get 156 from $S_{k+1}$, we would have to remove a term equal to $t_{k+1} + R$; but that's impossible since there is no term greater than $t_{k+1}$, the last term in the series. By the same logic, adding more terms to the series will keep pushing us further from a solution.

So the only possible solution would happen for the first $n$ for which $n(2n+1) > 156$, i.e. $n=9$. Happily, $S_9=171$, and the difference $15$ is the fourth term in your series.