The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is four times its 15th term.

We know that

a_n = a + (n - 1)d

Given,
a₂₄ = 2(a₁₀)
∴ a + 23d = 2(a + 9d)
⇒ a + 23d = 2a + 18d
⇒ 2a - a = 23d - 18d
⇒ a = 5d.

By using formula for nth term and a = 5d

a₇₂ = a + 71d = 5d + 71d = 76d
a₁₅ = a + 14d = 5d + 14d = 19d

∴ 4a₁₅ = 4 × 19d = 76d = a₇₂.

Hence, proved that 72nd term is four times the 15th term of the A.P.