# In an arithmetic progression, S_m = n and S_n = m. Also, m>n. What is the sum of the first (m - n) terms?

Dec 3, 2016

${\sum}_{i = 1}^{m - n} {S}_{i} = \frac{\left(m - n\right) \left(m + 3 n - 1\right)}{2}$

#### Explanation:

In an arithmetic progression with common difference $d$ and initial term ${a}_{1}$, we can write the ${k}^{\text{th}}$ term as

${a}_{k} = {a}_{1} + \left(k - 1\right) d$

In our given sequence, we can write ${S}_{m}$ and ${S}_{n}$ in the above form to get

$\left\{\begin{matrix}{S}_{m} = {S}_{1} + \left(m - 1\right) d = n \\ {S}_{n} = {S}_{1} + \left(n - 1\right) d = m\end{matrix}\right.$

$\implies \left({S}_{1} + \left(m - 1\right) d\right) - \left({S}_{1} + \left(n - 1\right) d\right) = n - m$

$\implies \left(m - n\right) d = n - m$

$\implies d = \frac{n - m}{m - n} = - 1$

Additionally, we can substitute that back into the second equation to get

${S}_{1} - n + 1 = m$

$\implies {S}_{1} = m + n - 1$

Next, the sum of the first $k$ terms of an arithmetic progression is given by

${\sum}_{i = 1}^{k} {a}_{i} = \frac{k \left({a}_{1} + {a}_{k}\right)}{2}$

so the sum of the first $m - n$ term of the given progression is

${\sum}_{i = 1}^{m - n} {S}_{i} = \frac{\left(m - n\right) \left({S}_{1} + {S}_{m - n}\right)}{2}$

$= \frac{\left(m - n\right) \left({S}_{1} + {S}_{1} + \left(m - n - 1\right) d\right)}{2}$

$= \frac{\left(m - n\right) \left(2 {S}_{1} - m + n + 1\right)}{2}$

$= \frac{\left(m - n\right) \left(2 \left(m + n - 1\right) - m + n + 1\right)}{2}$

$= \frac{\left(m - n\right) \left(m + 3 n - 1\right)}{2}$