If p times the pth term of an AP is an equal to q times the q th term. Find the (p+q)th term of the AP

p * ap = q * aq

p{ a + (p-1) d } = q { a + ( q - 1 ) d }

ap + p2d - pd = aq + q2d - qd

ap - aq = - p2d + q2d - qd + pd

a (p - q ) = d ( q2 - p2 + p - q )

a ( p - q ) = d { ( q - p ) ( q + p) + p - q }

a ( p - q ) = d ( p - q ) { -1 ( p + q) + 1 }

a = d ( - p - q + 1 ) .... ( 1)

( p + q ) th term = a + (n - 1 ) d

here , n = no. of terms

= a + ( p + q  - 1 ) d = o .. (2)

substituting a = d ( - p - q + 1 ) in ( 2 )

= d ( - q - p + 1 ) + ( p + q - 1 ) d

= -dp - dq + d + pd + qd - d

= 0