Given: (m + 1)^th term of an A.P. is twice the (n + 1)^th term

⇒ a_m+1 = 2a_n+1

To prove: a_3m+1 = 2a_m+n+1

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 = m + 1:

∴ a_m+1 = a + (m + 1 – 1)d

⇒ a_m+1 = a + md

When n = n + 1:

∴ a_n+1 = a + (n + 1 – 1)d

⇒ a_n+1 = a + nd

According to question:

a_m+1 = 2a_n+1

⇒ a + md = 2(a + nd)

⇒ a + md = 2a + 2nd

⇒ a – 2a + md – 2nd = 0

⇒ -a + (m – 2n)d = 0

⇒ a = (m – 2n)d………(i)

a_n = a + (n – 1)d

When n = m + n + 1:

∴ a_m+n+1 = a + (m + n + 1 – 1)d

⇒ a_m+n+1 = a + md + nd

⇒ a_m+n+1 = (m – 2n)d + md + nd……… (From (i))

⇒ a_m+n+1 = md – 2nd + md + nd

⇒ a_m+n+1 = 2md – nd………(ii)

When n = 3m + 1:

∴ a_3m+1 = a + (3m + 1 – 1)d

⇒ a_3m+1 = a + 3md

⇒ a_3m+1 = (m – 2n)d + 3md……… (From (i))

⇒ a_3m+1 = md – 2nd + 3md

⇒ a_3m+1 = 4md – 2nd

⇒ a_3m+1 = 2(2md – nd)

⇒ a_3m+1 = 2a_m+n+1…………(From (ii))

Hence Proved