Given: \(\frac{a}{1-r} = 20\) & \(\frac{a^2}{1-r^2}\) = 100
(Because on squaring both first term a and common ratio r will be squared.)
To find: the series
a = 20(1-r) ...... (1)
Put this value of r in equation (i) we get
∴The infinite geometric series is:8, \(\frac{24}{5}, \frac{72}{25}, \frac{216}{125}, \frac{648}{625}, .... \infty\)
