If the $ 9^{\text {th }} $ term of an AP is zero, prove that its $ 29^{\text {th }} $ term is twice its $ 19^{\text {th }} $ term.

Given: 

The \( 9^{\text {th }} \) term of an AP is zero.

To do: 

We have to prove that its \( 29^{\text {th }} \) term is twice its \( 19^{\text {th }} \) term.

Solution:

Let $a$ be the first term and $d$ be the common difference.

This implies,

$a_{9}=a+(9-1)d$

$0=a+8d$

$a=-8d$........(i)

$a_{29}=a+(29-1)d$

$=a+28d$

$=-8d+28d$            [From (i)]

$=20d$.........(ii)

$a_{19}=a+(19-1)d$

$=a+18d$

$=-8d+18d$            [From (i)]

$=10d$........(iii)

Therefore,

$a_{29}=2(10d)$

$=2\times a_{19}$      [From (iii)]

Hence proved.