If 9th term of an A.P. is Zero, prove that its 29th term is double the 19th term.

Answer:

Given,

9^th term of an A.P is 0

a₉ = 0

a_n = a + (n – 1) d [w

When n = 9,

a₉ = a + (9 – 1)d

= a + 8d

 

a₉ = 0

a + 8d = 0

a = -8d

When n = 19,

a₁₉ = a + (19 – 1)d

= a + 18d

= -8d + 18d

= 10d

When n = 29,

a₂₉ = a + (29 – 1)d

= a + 28d

= -8d + 28d [Since, a = -8d]

= 20d

= 2×10d

a₂₉ = 2a₁₉

Thus, Proved.