In an A.P. ten times of its tenth term is equal to thirty times of its 30^th term. Find its 40^th term.

Let the first term of the A.P. be a and it's common difference be d.

According to question,

⇒ 10a₁₀ = 30a₃₀

⇒ 10[a + (10 - 1)d] = 30[a + (30 - 1)d]

⇒ a + 9d = 3(a + 29d)

⇒ a + 9d = 3a + 87d

⇒ 3a - a = 9d - 87d

⇒ 2a = -78d

⇒ a = -39d.

∴ a₄₀ = a + (40 - 1)d

= -39d + 39d

= 0.

Hence, a₄₀ = 0.