If 8th term of an A.P. is zero, prove that its 38th term is triple of its 18th term.

Given,
a₈ = 0

By using formula a_n = a + (n - 1)d for a₈ we get,

⇒ a + (8 - 1)d = 0
⇒ a + 7d = 0
⇒ a = -7d.

So,
a₃₈ = a + (38 - 1)d
      = a + 37d
      = -7d + 37d
      = 30d.

a₁₈ = a + (18 - 1)d
      = a + 17d
      = -7d + 17d
      = 10d.

∴ a₃₈ = 30d = 3 × 10d = 3 × a₁₈.

Hence, proved that 38th term of the A.P. is triple of the 18th term.