If (m + l)th term of an A.P. is twice the (n + l)th term. Prove that (3m + l)th term is twice the (m + n + l)th term.
Let a be the first and d be the common difference of an A.P. Since (m + 1)th term of an A.P. is twice the (n + 1)th term.
∴
or a + (m + 1 - 1)d = 2[a + (n + 1 - 1)d]
or a + md = 2a + 2nd
or a = (m - 2n)d
Now,
...(i)
Now,
or ...(ii)
∴ (3m + 1)th term is twice the (m + n + 1)th term of an A.P.