Answer
$a_n=2 (\dfrac{2}{3})^{n-1}$
$a_{15}=\dfrac{32768}{4,782, 969}$
$S_{15}= 5.99$
Work Step by Step
The general formula for the nth term of a geometric series is given by $a_n= a_1r^{n-1}$ ...(1)
The ratio of the successive terms is $\dfrac{2}{3}$.
Equation (1) gives: $a_n= a_1 \times \dfrac{2}{3}^{n-1}=2 (\dfrac{2}{3})^{n-1}$
Plugging in $n =15$, we have
$a_{15}=2 (\dfrac{2}{3})^{15-1}=2 (\dfrac{2}{3})^{14}=\dfrac{32768}{4,782, 969}$
We know that $S_{n}=\dfrac{a_1(1-r^n)}{1-r}$
Now, $S_{15}=\dfrac{2(1-\dfrac{2}{3}^{15})}{1-\dfrac{2}{3}} = 5.99$
