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 `a_(3m + 1) = 2a_(m + n +1)`

So, let us first find the two terms.

As we know,

`a_n = a + (n' - 1)d`

For (m + 1)^th term (n’ = m + 1)

`a_(m + 1) = a + (m + 1 - 1)d`

= a + md

For (n + 1)^th term (n’ = n + 1),

`a_(n +1) = a + (n + 1 -1)d`

= a + nd

Now, we are given that `a_(m + 1) = 2a_(n +1)`

So we get

a + md = 2(a + nd)

a + md = 2a + 2nd

md - 2nd = 2a - a

(m - 2n)d = a  ...........(1)

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)

`a_(m + n + 1) = a + (m +n +1 -1)d`

= (m - 2n)d + (m + n)d

= md - 2nd + md + nd   (Using 1)

= 2md - nd

For (3m + 1)^th term (n’ = 3m + 1),

`a_(3m +1) = a + (3m + 1 -1)d`

= (m - 2n)d + 3md        (using 1)

= md - 2nd + 3md

= 4md - 2nd

= 2(2md - nd)

Therefore `a_(3m + 1) = 2a_(m + n + 1)`

Hence proved