Precalculus (6th Edition) Blitzer

$\frac{1}{2}$,$\frac{6}{7}$ , $\frac{9}{8}$, $\frac{4}{3}$
To ﬁnd the first four terms of the sequence whose general term is $a_{n}$ = $\frac{3n}{n+5}$, we replace n in the formula with 1,2,3, and 4. n=1, $a_{1}$ = $\frac{3n}{n+5}$ =$\frac{3*1}{1+5}$ = $\frac{3}{6}$ = $\frac{1}{2}$ n=2, $a_{2}$ =$\frac{3n}{n+5}$ = $\frac{3*2}{2+5}$ = $\frac{6}{7}$ = $\frac{6}{7}$ n=3, $a_{3}$ = $\frac{3n}{n+5}$ =$\frac{3*3}{3+5}$ = $\frac{9}{8}$ n=4,$a_{4}$ = $\frac{3n}{n+5}$ = $\frac{3*4}{4+5}$ = $\frac{12}{9}$ =$\frac{4}{3}$ The ﬁrst four terms are $\frac{1}{2}$,$\frac{6}{7}$ , $\frac{9}{8}$, $\frac{4}{3}$.
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