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

##### 1 Answer

#### Explanation:

In an arithmetic progression with common difference

In our given sequence, we can write

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

Next, the sum of the first

so the sum of the first

#=((m-n)(S_1+S_1+(m-n-1)d))/2#

#=((m-n)(2S_1-m+n+1))/2#

#=((m-n)(2(m+n-1)-m+n+1))/2#

#=((m-n)(m+3n-1))/2#