Given: 24^th term is twice the 10^th term

⇒ a₂₄ = 2a₁₀

To prove: a₇₂ = 2a₃₄

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

When n = 10:

∴ a₁₀ = a + (10 – 1)d

⇒ a₁₀ = a + 9d

When n = 24:

∴ a₂₄ = a + (24 – 1)d

⇒ a₂₄ = a + 23d

When n = 34:

∴ a₃₄ = a + (34 – 1)d

⇒ a₃₄ = a + 33d ………(i)

When n = 72:

∴ a₇₂ = a + (72 – 1)d

⇒ a₇₂ = a + 71d

According to question:

a₂₄ = 2a₁₀

⇒ a + 23d = 2(a + 9d)

⇒ a + 23d = 2a + 18d

⇒ a – 2a + 23d – 18d = 0

⇒ -a + 5d = 0

⇒ a = 5d

Now, a₇₂ = a + 71d

⇒ a₇₂ = 5d + 71d

⇒ a₇₂ = 76d

⇒ a₇₂ = 10d + 66d

⇒ a₇₂ = 2(5d + 33d)

{∵ a = 5d}

⇒ a₇₂ = 2(a + 33d)

⇒ a₇₂ = 2a₃₄ (From (i))

Hence Proved