Given,

10 times the 10^th term of an A.P. is equal to 15 times the 15^th term

⇒ 10a₁₀ = 15a₁₅

To prove: a₂₅ = 0

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 = 15:

∴ a₁₅ = a + (15 – 1)d

⇒ a₁₅ = a + 14d

When n = 25:

∴ a₂₅ = a + (25 – 1)d

⇒ a₂₅ = a + 24d ………(i)

According to question:

10a₁₀ = 15a₁₅

⇒ 10(a + 9d) = 15(a + 14d)

⇒ 10a + 90d = 15a + 210d

⇒ 10a – 15a + 90d – 210d = 0

⇒ -5a – 120d = 0

⇒ -5(a + 24d) = 0

⇒ a + 24d = 0

⇒ a₂₅ = 0 (From (i))

Hence Proved