If the 9th term of an AP is 0, prove that its 29th term is twice its 19th term.

Let a and d respectively be the first term and common difference of the AP.

Given a₉ = 0

So, a + (9-1)d = 0

a+8d=0

a= -8d

 

Now, 29th term = a+28d

=-8d+28d

= 20d = 2 x 10d

= 2(-8d + 18d)

=2(a+18d)

= 2 x 19th term

 

Thus, the 29th term of the AP is twice the 19th term.