In an AP, Sp = q, Sq = p and Sr denotes the sum of first r terms. Then, Sp + q is equal to

Sp = {2A + (p–1) D} = q, given


Sq = (2A + (q–1) D} = p, given


On Subtracting the second equation from 1st we get,


D = –2


Also On adding the two equations we get,


2A + = p/q + q/p


Now,


Sp + q = {2A + (p + q–1) D}


= {2A + (p + q–1)D}


{ + + (p + q–1) D}


=


[ By substituting the value of D ]


= – (p + q)

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