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

Question

2350 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-03T15:34:07

#sum_(i=1)^(m-n)S_i=((m-n)(m+3n-1))/2#

In an arithmetic progression with common difference #d# and initial term #a_1#, we can write the #k^"th"# term as

#a_k = a_1+(k-1)d#

In our given sequence, we can write #S_m# and #S_n# in the above form to get

#{(S_m=S_1+(m-1)d = n),(S_n=S_1+(n-1)d=m):}#

#=> (S_1+(m-1)d) - (S_1+(n-1)d) = n-m#

#=> (m-n)d = n-m#

#=> d = (n-m)/(m-n) = -1#

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

#S_1-n+1 = m#

#=> 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 = (k(a_1+a_k))/2#

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

#sum_(i=1)^(m-n)S_i = ((m-n)(S_1+S_(m-n)))/2#

#=((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#