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

1.	If (m+1)th term of an A.P. is twice the (n+1)th term, prove that (3m+1)th term is twice (m+n+1)th term.
Answer» 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: a_3m+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

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

Answer»

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:

a_3m+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