Answer
$\dfrac{1}{2},\dfrac{6}{7},\dfrac{9}{8},\dfrac{12}{9}$
Work Step by Step
Need to find the first four terms of $a_n=\dfrac{3n}{n+5}$ when $n=1,2,3,4$
Thus,
$a_1=\dfrac{3(1)}{1+5}=\dfrac{3}{6}=\dfrac{1}{2}$
$a_2=\dfrac{3(2)}{2+5}=\dfrac{6}{7}$
$a_3=\dfrac{3(3)}{3+5}=\dfrac{9}{8}$
$a_4=\dfrac{3(4)}{4+5}=\dfrac{12}{9}$
Hence, the first four terms are: $\dfrac{1}{2},\dfrac{6}{7},\dfrac{9}{8},\dfrac{12}{9}$
