#### Question

In a certain A.P. the 24^{th} term is twice the 10^{th} term. Prove that the 72nd term is twice the 34th term.

#### Solution

Here, we are given that 24^{th} term is twice the 10^{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 have to prove that `a_72 = 2a_34`

So, let us first find the two terms.

As we know

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

For 10th term (n = 10)

`a_10 = a + (10 - 1)d`

= a + 9d

For 24 th term (n = 24)

`a_24 = a + (24 - 1)d`

= a + 23d

Now we given that `a_24 = 2a_10`

So we get

a + 23d = 2(a + 9d)

a + 23d = 2a + 18d

23d - 18d = 2a - a

5d = a .....(1)

Further, we need to prove that the 72^{nd} term is twice of 34^{th} term. So let now find these two terms,

For 34^{th} term (n = 34),

`a_34 = a + (34 - 1)d`

= 5d + 33d (Using 1)

= 38d

For 72^{nd} term (n = 72)

`a_72 = a + (72 - 1)d`

= 5d + 71d (using 1)

= 76d

= 2(38d)

Therefore `a_72 = 2a_34`