Question

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

Answer 1

Given:

In an AP, the (p + 1)th term is twice the (q + 1)th term.

`a_(p + 1) = 2a_(q + 1)`

To prove: `a_(3p + 1) = 2a_(p + q + 1)`

Use the nth term formula of an A.P.

a_n = a + (n − 1)d

`a_(p + 1) = a + pd`

`a_(q + 1) = a + qd`

Use the given condition

a + pd = 2(a + qd)

a + pd = 2a + 2qd

pd − 2qd = 2a − a

(P − 2q)d = a   ...(1)

Find the (3p + 1)th term

`a_(3p + 1) = a + 3pd`

Substitute aaa from (1)

`a_(3p + 1) = (p − 2q)d + 3pd`

= (4p − 2q)d

= 2(2p − q)d   ...(2)

Find the (p + q + 1)th term

`a_(p + q + 1) = a + (p + q)d`

Substitute aaa from (1)

`a_(p + q + 1) = (p + 2q)d + (p + q)d`

= (2p + q)d   ...(3)

From (2) and (3):

`a_(3p + 1)​ = 2(2p − q)d = 2a_(p + q + 1)​`

Hence proved `a_(3p + 1)​ = 2a_(p + q + 1)​`