Prove that: 29^th term is double the 19^th term (i.e. a₂₉ = 2a₁₉)

Given: a₉= 0

(Where a=a₁ is first term, a₂ is second term, a_n is nth term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

So a₉= 0 a + (9 - 1)d = 0

a + 8d = 0

a = (–8d) ….equation (i)

Now a₂₉ = a + (29 - 1)d and a₁₉ = a + (19 - 1)d

a₂₉= a + 28d and a₁₉ = a + 18d ….equation (ii)

By using equation (i) in equation (ii), we have

a₂₉= –8d + 28d and a₁₉ = –8d + 18d

a₂₉= 20d and a₁₉ = 10d

So a₂₉= 2a₁₉

HENCE PROVED