Solve this

Solution:

Prove that: $29^{\text {th }}$ term is double the $19^{\text {th }}$ term (i.e. $a_{29}=2 a_{19}$ )

Given: $a_{9}=0$

(Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given $\mathrm{AP}$ )

Formula Used: $a_{n}=a+(n-1) d$

So $a_{9}=0 \rightarrow a+(9-1) d=0$

$a+8 d=0$

$\mathrm{a}=(-8 \mathrm{~d})$....equation (i)

Now $\mathrm{a}_{29}=\mathrm{a}+(29-1) \mathrm{d}$ and $\mathrm{a}_{19}=\mathrm{a}+(19-1) \mathrm{d}$

$a_{29}=a+28 d$ and $a_{19}=a+18 d \ldots$ equation (ii)

By using equation (i) in equation (ii), we have

$a_{29}=-8 d+28 d$ and $a_{19}=-8 d+18 d$

$a_{29}=20 d$ and $a_{19}=10 d$

$\underline{\text { So }} a_{29}=2 a_{19}$

HENCE PROVED