Answer:
First Term = 6
Common Ratio = 3
Step-by-step explanation:
According to the Question,
Thus, [tex]S_{8} = 19680[/tex] & [tex]S_{4} = 240[/tex] .
Now, on solving [tex]\frac{S_{8} }{S_{4} }[/tex] we get,
[tex]\frac{19680}{240} = \frac{\frac{a(r^{8-1)} }{r-1}}{\frac{a(r^{4-1)} }{r-1}}[/tex]
[tex]82 = \frac{r^{8}-1 }{r^{4}-1 }[/tex]
[tex]82r^{4}-82 = r^{8}-1\\r^{8}-82r^{4}+81 = 0\\r^{8}-81r^{4}-r^{4}+81 = 0\\(r^{4}-81)( r^{4}-1) =0[/tex](r=1 is not possible so neglect [tex]( r^{4}-1) =0[/tex] )
So, r=3 Now Put this value in [tex]S_{4} = {\frac{a(r^{4-1)} }{r-1}}[/tex] We get a=6 .