Given:
(m+1)th term of an A.P. is twice the (n+1)th term a(m +1) = 2a(n +1)
To Prove:
(3m+1)th term is twice (m+n+1)th term a(3m + 1) = 2a(m +n +1)
Proof:
a(m +1) = 2a(n +1)
⇒ a + (m + 1 – 1) d = 2a + 2(n +1 -1)d
⇒ - a = 2nd – md
⇒ a = md - 2nd (i)
LHS:
a3m+1 = a + (3m + 1 -1)d = md – 2nd + 3md = 2d(2m - n)
RHS:
2a(m + n + 1) = 2[a +(m +n +1 -1)d] = 2[md - 2nd + md + nd] = 2d(2m - n)
LHS = RHS
Hence, proved