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)nth terms is (m/2)(mn-m+n+1)(m2)(mnm+n+1)?

1 Answer
Aug 28, 2017

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

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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)