If the $9^{th}$ term of Arithmetic Progression (AP) is equal to zero, then $29^{th}$ term of AP is twice the term:

Question

If the $9^{th}$ term of Arithmetic Progression (AP) is equal to zero, then $29^{th}$ term of AP is twice the term:

$a$. $11^{th}$

$b$. $13^{th}$

$c$. $19^{th}$

$d$. $14^{th}$.

My Attempt:

$$t_{9}=0$$ $$a+8d=0$$

Where $a$ is first term and $d$ is common difference.

Now, $t_{29}=a+28d$

How do I proceed further?

Answer 1

\begin{align}t_{29}&=a+28d\\&=a+8d+20d\\&=20d\\&=2(10d) \\&=2(a+8d+10d)\\&=2(a+18d) \end{align}

Answer 2

Hint:

$a+8d=0$. So $t_{29}=20d$.

Which term is $10d$?

Answer 3

So we have $t_{29}=t_9+20d$

If $t_9=0$, then $t_{29} = 2(10d)$.

So half of $t_{29}$ is $10d$, which is equal to $t_9 + 10d$, or $t_{19}$ (C)

Answer 4

Let the given AP be {$T_i$}.

Consider a new AP {$U_j$} where $U_j=T_{j-9}$, i.e. $U_0=T_9=0$.

If follows that $U_1=a, U_2=2a, U_3=3a, \cdots, U_j=ja$, i.e. $U_j\propto j$

Hence $$\color{red}{T_{29}}=U_{20}=2U_{10}\color{red}{=2T_{19}}$$