## College Algebra (6th Edition)

The Sum of the given infinite geometric series = $\frac{9}{4}$
The Sum of a infinite geometric series ( if | r | $\lt$ 1 ) is given by S = $\frac{First term }{1 - common ratio}$ = $\frac{a_{1}}{1 - r}$ The given infinite geometric series = 3 - 1 + $\frac{1}{3}$ - $\frac{1}{9}$ + ...................... Here First term $a_{1}$ = 3 Common ratio r = $\frac{-\frac{1}{9}}{\frac{1}{3}}$ = $\frac{\frac{1}{3}}{-1}$ = $\frac{-1}{3}$ = - $\frac{1}{3}$ The Sum of the given infinite geometric series = $\frac{a_{1}}{1 - r}$ = $\frac{3}{1 - (-\frac{1}{3})}$ = $\frac{3}{1 + \frac{1}{3}}$ = $\frac{3}{\frac{4}{3}}$ = 3$\times$$\frac{3}{4}$ = $\frac{9}{4}$