$a_{n+1}=5 ^n a_n$

the above terms can be give as $\begin{aligned} a_1&=5^0 \\ a_2&=5^{0+1} \\ a_3&=5^{0+1+2} \\ a_4&=5^{0+1+2+3} \\ &\vdots \\ a_n&=5^{n(n-1)/2}=5^{276} \end{aligned}$

• we know that $1+2+3+4...n= n(n+1)/2$
• here, $2n(n+1)/2=276$ which gives $n=23$
• $k=n+1=24$

Let $b_n = \log_5 a_n$ . Then $b_1 = \log_5 a_1 = \log_5 1 = 0$ and $b_{n+1} = \log_5 \left(5^na_n\right) = n + b_n$ . Then we have:

$\begin{aligned} b_{n+1} - b_n & = n \\ \sum_{n=1}^k b_{n+1} - \sum_{n=1}^k b_n & = \sum_{n=1}^k n \\ b_{k+1} - \color{#3D99F6} b_1 & = \frac {k(k+1)}2 & \small \color{#3D99F6} \text{Since }b_1 = 0 \\ b_{k+1} & = \frac {k(k+1)}2 & \small \color{#3D99F6} \text{Replace }k \text{ with }k-1. \\ \implies b_k & = \frac {k(k-1)}2 \end{aligned}$

When $a_k = 5^{276} \implies b_k = 276$ , $\implies \dfrac {k(k-1)}2 = 276$ $\implies k = \left \lceil \sqrt{2\times 276}\right \rceil = \boxed{24}$ .