Question

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

Answer 1

Let $a,d$ - first term and the common difference of A.P

$n^{th}$ term , $a_n=a+(n-1)d$

$\Rightarrow a_9=a+8d=0\Rightarrow a=-8d$

$\frac{a_{29}}{a_{19}}=\frac{a+28d}{a+18d}=\frac{-8d+28d}{-8d+18d}=\frac{20d}{10d}=2$

$\therefore$ $a_{29}=2a_{19}$