Let a_1=24 and form the sequence a_n, n ≥ 2 by a_n=100 a_ n-1 +134...

$a_1=24 \& a_n=100 . a_{n-1}+134$ First few terms : $$\begin{aligned} & a_1=24 \\ & a_2=2534 \end{aligned}$$ $$\begin{aligned} & a_3=253534 \\ & a_4=25353534 \\ & \therefore a_n=2 \underbrace{535353 \ldots \ldots . .534}_{(n-1) \text { Times }^{\prime} 53^{\prime}}\end{aligned}$$ Now, an $\rightarrow$ divisible by $99 \Rightarrow$ by $9 \& 11$ both Sum of digits $=6+8(n-1)$ To be divisible by 9 , $n=7,16,25,34,43,52,61,70,79,88, \ldots .$ $\mathrm{a}_7=2 \underbrace{535353535353}_{6 \text { Times }{ }^{\prime} 53^{\prime}} 4$ But $\mathrm{a}_7 \rightarrow$ Not divisible by 11 . $\mathrm{a}_{16}=2 \underbrace{5353535353 \ldots . .53}_{15 \text { Times'53' }} 4$ Similarly, $\mathrm{a}_{16} \rightarrow$ Not divisible by 11 . Now, $\mathrm{n}=88$ $\mathrm{a}_{88}=2 \underbrace{5353 \ldots \ldots 53}_{87 \text { Times }{ }^{\prime} 53^{\prime}} 4$ \begin{aligned} & \text { Divisibility by } \begin{aligned} 11 & \rightarrow|(2+3+3 \ldots . .)-(5+5+\ldots .4)| \\ & =|263-439| \\ & =176\end{aligned} \\ & \therefore \text { Least } n=88\end{aligned}$