Let a be the first term and d be the common difference.

Given: a₂₄ = 2(a₁₀)

To prove: a₇₂ = 4 × a₁₅

Now, Consider a₂₄ = 2a₁₀

⇒ a + 23d = 2[a + 9d]

⇒ a + 23d = 2a + 18d

⇒ a = 5d …………………. (1)

Consider a₇₂ = a + (72 - 1)d

⇒ a₇₂ = 5d + 71d (from equation (1))

⇒ a₇₂ = 76d ………………. (2)

Now, consider a₁₅ = a + (15 - 1)d

⇒ a₁₅ =5d + 14d (from equation (1))

⇒ a₁₈ = 19d ………………….(3)

From equation (2) and (3), we get,

a₇₂ = 4 × a₁₅

Hence, proved.