## Calculus: Early Transcendentals 8th Edition

converges and its sum is $\frac{400}{9}$
In this explicit form we see that $r$ =|$0.73|\lt1$ which means that this geometric series will converge. Knowing that $a$ refers to the first term which is $12$ and using the explicit $r$, we get $S= \frac{a}{1-r}$ = $\frac{12}{1-0.73} = \frac{12}{0.27} = \frac{400}{9}$ Hence, the series converges and its sum is $\frac{400}{9}$ .