## Algebra 2 (1st Edition)

$a_1=\dfrac{100}{31}\\a_2=\dfrac{200}{31} \\ a_3=\dfrac{400}{31}\\ a_4=\dfrac{800}{31} \\a_5=\dfrac{1600}{31}$
We know that $S_{n}=a_1(\dfrac{1-r^{n}}{1-r})$ Here, $a_1=2(3)^{1-1}=2$ and $r=3$ Now, $S_{5}=a_1(\dfrac{1-r^{5}}{1-r})$ or, $100=a_1(\dfrac{1-r^{5}}{1-r})$ or, $100=a_1(\dfrac{1-2^{5}}{1-2})$ or, $100=a_1(\dfrac{1-2^{5}}{-1}) \implies a_1=\dfrac{100}{31}\\a_2=\dfrac{100}{31} \times 2=\dfrac{200}{31} \\ a_3=\dfrac{200}{31} \times 2=\dfrac{400}{31}\\ a_4=\dfrac{400}{31} \times 2=\dfrac{800}{31} \\a_5=\dfrac{800}{31} \times 2=\dfrac{1600}{31}$