Chapter 12 Sequences and Series - 12.4 Find Sums of Infinite Geometric Series - 12.4 Exercises - Mixed Review - Page 825: 55

$a_n=\dfrac{1}{9} (6)^{n-1}$

Here, we have $a_n= a_1r^{(n-1)}$ for the Geometric series. We are told that $a_4=24$ and $r=6$ Here, we have $a_4=(a_1) \times (6)^{(4-1)} \implies 24=(a_1) \times (6)^3$ or, $a_1=\dfrac{24}{216}=\dfrac{1}{9}$ Now, $a_n=a_1 \times r^{n-1}$ Hence, $a_n=\dfrac{1}{9} (6)^{n-1}$