If (m + 1)th term of an A.P. is twice the (n + 1)th term, then prove t

1.	If (m + 1)th term of an A.P. is twice the (n + 1)th term, then prove that (3m + 1)th term is twice the (m + n + 1)th term.
Answer» t_n = a + (n – 1)d Given t_{m + 1} = 2 t_{n + 1} a + (m + 1 – 1)d = 2[a + (n + 1 – 1)d] a + md = 2(a + nd) ⇒ a + md = 2a + 2nd md – 2nd = a d(m – 2n) = a … (1) To Prove t_{(3m + 1)} = 2(t_{m + n + 1}) L.H.S. = t_{3m + 1} = a + (3m + 1 – 1)d = a + 3md = d(m – 2n) + 3md (from 1) = md – 2nd + 3md = 4md – 2nd = 2d (2m – n) R.H.S. = 2(t_{m + n + 1}) = 2 [a + (m + n + 1 – 1)d] = 2 [a + (m + n)d] = 2 [d (m – 2n) + md + nd)] (from 1) = 2 [dm – 2nd + md + nd] = 2 [2 md – nd] = 2d (2m – n) R.H.S = L.H.S ∴ t_{(3m + 1)} = 2 t_{(m + n + 1)} Hence it is proved.

If (m + 1)th term of an A.P. is twice the (n + 1)th term, then prove that (3m + 1)th term is twice the (m + n + 1)th term.

Answer»

t_n = a + (n – 1)d

Given t_{m + 1} = 2 t_{n + 1}

a + (m + 1 – 1)d = 2[a + (n + 1 – 1)d]

a + md = 2(a + nd) ⇒ a + md = 2a + 2nd

md – 2nd = a

d(m – 2n) = a … (1)

To Prove t_{(3m + 1)} = 2(t_{m + n + 1})

L.H.S. = t_{3m + 1}

= a + (3m + 1 – 1)d

= a + 3md

= d(m – 2n) + 3md (from 1)

= md – 2nd + 3md

= 4md – 2nd

= 2d (2m – n)

R.H.S. = 2(t_{m + n + 1})

= 2 [a + (m + n + 1 – 1)d]

= 2 [a + (m + n)d]

= 2 [d (m – 2n) + md + nd)] (from 1)

= 2 [dm – 2nd + md + nd]

= 2 [2 md – nd]

= 2d (2m – n)

R.H.S = L.H.S

∴ t_{(3m + 1)} = 2 t_{(m + n + 1)}

Hence it is proved.