If the 9th term of an A.P. is zero, then prove that 29th term is double of 19th term.

#### Solution 1

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

9th term i.e n = 9

`:. t_9 = a+(9 -1)d`

= a + 8d

It is given that `t_9 = 0`

∴ a + 8d = 0 ....(i)

29th term i.e `t_29` where n = 29

`:. t_29 = a + (29-1)d`

`t_29 = a + 28d` ....(ii)

= (a + 8d) + 20d

= 0 + 20d ...by equ (i)

`:. t_29 = 20d` ....(ii)

`t_29 = a + (19 - 1)d`

`t_19` = a + 18d

= a + 8d + 10d

= 0 + 10d

`t_19 = 10d` .....(iiii)

by equation (ii) & (iii)

`t_29 = 2t_19`

#### Solution 2

In the given problem, the 9^{th} term of an A.P. is zero.

Here, let us take the first term of the A.P as *a* and the common difference as *d*

So, as we know,

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

We get

`a_9 = a + (9 - 1)d`

0 = a + 8d

a = -8d .......(1)

Now, we need to prove that 29^{th} term is double of 19^{th} term. So, let us first find the two terms.

For 19th term (n = 19)

`a_19 = a + (19 - 1)d`

= -8d + 18d (using 1)

= 10d

For 29th term (n = 29)

`a_29 = a + (29 - 1)d`

`= - 8d + 28d`

= 20d

= 2 x 10d

`= 2 x a_19`

(Using 1)

Therefore for the given A.P. the 29^{th} term is double of the 19^{th} term.

Hence proved.