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The sum of $m$ terms of an arithmetic series is $n$, and that of $n$ terms is $m$. Then how do we calculate the sum of $m+n$ terms?

We know this:

  • The sum of $p$ terms of an arithmetic series is $\frac{p}{2}(2a+(p-1)d)$ where $a$ is the first term and $d$ is the difference between each term.
  • We can express what $m$ and $n$ equal to by putting $p$ equal to $n$ and $m$ respectively.
  • Then to get $m+n$, we simply add the new ways of expressing $m$ and $n$.

Now to get the sum, we take $p=m+n$, but that yields an expression that is way too big to handle, and since this is a textbook problem, I am assuming the answer is short and succint.

Also, can you guys actually show me the example of a series where sum of $m$ terms is $n$ and vice versa?

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2 Answers 2

up vote 4 down vote accepted

You know that

$$na+\frac{n(n-1)}{2}d=m$$ $$ma+\frac{m(m-1)}{2}d=n$$

This is a system of two equations with unknowns $a$ and $n$. Multilying the first equation by $m$ and second by $n$, and subtracting, you get

$$mn\frac{n-m}{2}d = m^2-n^2$$

Thus

$$d= \frac{2(m-n)(m+n)}{mn(n-m)}$$

Cancel $m-n$ and you found$d$, any equation will yield $a$.

Note that the solution only works if $m \neq n$. If $m=n$, the problem doesn't have unique solution, actually any number can be the answer in that case.

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@rahul Typo, fixed ty. –  N. S. Aug 16 '13 at 18:11

For an example take the arithmetic progression $2,-1, -4, \dots$ with difference $d=-3$

The sum of one term is $2$ and the sum of two terms is $1$

Or $\frac 73,\frac23,-1 \dots $ where the sum of two terms is $3$ and the sum of three terms is $2$

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Hmm..I made a premature assumption that sum of any m terms is equal to n and that of any n is m. –  rah4927 Aug 17 '13 at 16:06

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