Answer
a) $2 +2(\dfrac{3}{5})+2(\dfrac{3}{5})^2+....$ ;
b) $\dfrac{13}{2}-\dfrac{13}{2}(\dfrac{3}{10})+\dfrac{13}{2}(\dfrac{3}{10})^2+....$
Work Step by Step
a) Given: $\dfrac{2}{1-r}=5$
and $1-r=\dfrac{2}{5}$ or, $ r=\dfrac{3}{5}$
Hence, the series will be: $2 +2(\dfrac{3}{5})+2(\dfrac{3}{5})^2+....$
b) Given: $\dfrac{\dfrac{13}{2}}{1-r}=5$
and $1-r=\dfrac{13}{10}$ or, $ r=\dfrac{-3}{10}$
Hence, the series will be: $\dfrac{13}{2}-\dfrac{13}{2}(\dfrac{3}{10})+\dfrac{13}{2}(\dfrac{3}{10})^2+....$
Hence, a) $2 +2(\dfrac{3}{5})+2(\dfrac{3}{5})^2+....$ ;
b) $\dfrac{13}{2}-\dfrac{13}{2}(\dfrac{3}{10})+\dfrac{13}{2}(\dfrac{3}{10})^2+....$
