Answer
$a_1=5$
$a_2=\frac{17}{4}$
$a_3=\frac{7}{2}$
$a_4=\frac{11}{4}$
$a_5=2$
Work Step by Step
$a_n=a_1+(n-1)d$
$a_1=5,~~d=-\frac{3}{4}$
$a_n=5+(n-1)(-\frac{3}{4})$
$a_n=5-\frac{3}{4}n+\frac{3}{4}$
$a_n=-\frac{3}{4}n+(5+\frac{3}{4})=-\frac{3}{4}n+\frac{5(4)+3(1)}{4}$
$a_n=-\frac{3}{4}n+\frac{23}{4}$
$a_2=-\frac{3}{4}(2)+\frac{23}{4}=-\frac{6}{4}+\frac{23}{4}=\frac{17}{4}$
$a_3=-\frac{3}{4}(3)+\frac{23}{4}=-\frac{9}{4}+\frac{23}{4}=\frac{14}{4}=\frac{7}{2}$
$a_4=-\frac{3}{4}(4)+\frac{23}{4}=-\frac{12}{4}+\frac{23}{4}=\frac{11}{4}$
$a_5=-\frac{3}{4}(5)+\frac{23}{4}=-\frac{15}{4}+\frac{23}{4}=\frac{8}{4}=2$
