If there be 'm' A.P's beginning with unity whose common difference is 1,2,3......m. Show that the sum of their #n^(th)# terms is #(m/2)(mn-m+n+1)#?

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Answer 1 · 2017-08-28T09:35:58

The first AP series
#1,2,3....# which will have nth term #=n#

The 2nd AP series
#1,3,5,7....# which will have nth term #=1+(n-1)*2=2n-1#

The 3rd AP series
#1,4,9,12....# which will have nth term #=1+(n-1)*3=3n-2#

#------------#

The #m# th AP series
#1,(m+1),(2m+1),(3m+1)....# which will have nth term #=1+(n-1)*m=mn-m+1#

So sum of nth terms of m AP series will be

#S=m/2("first term"+"last term")#

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

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

Another approach

Here the number of terms for all nth terms of m AP series is #=m#

First term of the series #a=n#

The common difference of the series #d=(2n-1)-n=n-1#

So applying formula

#S=m/2[2n+(m-1)(n-1)]#

#=>S=m/2[2n+mn-n-m+1)]#

#=>S=m/2(mn+n-m+1)#