Question

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

Answer 1

Here, we are given that (m+1)^th term is twice the (n+1)^th term, for a certain A.P. Here, let us take the first term of the A.P. as a and the common difference as d

We need to prove that

So, let us first find the two terms.

As we know,

For (m+1)^th term (n’ = m+1)

For (n+1)^th term (n’ = n+1),

Now, we are given that

So, we get,

Further, we need to prove that the (3m+1)^th term is twice of (m+n+1)^th term. So let us now find these two terms,

For (m+n+1)^th term (n’ = m+n+1 ),

(Using 1)

For (3m+1)^th term (n’ = 3m+1),

Therefore,

Hence proved