Question

If p times the pth term of an A.P. is equal to q times its qth term, show that the (p + q)th term of the A.P. is zero.

Answer 1

Let a be the first term and d be the common difference of the given A.P.

Given: (p times pth term)  = (q times qth term)

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 }

[ a² - b² = (a+b)(a-b)]
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 = (p+q)
(p + q)th = a + ( p + q  - 1 ) d ………….(2)
substituting a = d ( - p - q + 1 ) in eq. ( 2 )
(p + q)th = d (- q - p + 1 ) + ( p + q - 1 ) d
= -dp - dq + d + pd + qd - d
(p + q)th = 0

Hence, (p + q)th term of an A.P is zero.

HOPE THIS WILL HELP YOU....