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

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

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

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

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)n^(th) terms is (m/2)(mn-m+n+1)(m/2)(mn-m+n+1)?