If the 9th term of an AP is zero, then prove that its 29th term is twice its 19th term.
Let the first term, common difference and number of terms of an AP are a, d and n respectively.
Given : 9th term is zero i.e. a9 = 0
To prove : a29 = 2a19
Proof :
As a9 = 0
a + 8d = 0 [ eqn i]
Using the nth term formula i.e. an = a + (n - 1)d
Taking LHS
a29 = a + 28d
= (2 - 1)a + (36 - 8)d
= 2a - a + 36d - 8d
= 2a + 36d - (a + 8d)
= 2(a + 18d) - 0 [ using i]
= 2a9 [ as a9 = a + 8d]
= RHS