In the certain geometrical sequence (an) a4 is eight times bigger than a1. a2 = 6 Find the smallest natural number k so that ak > 80
The nth term of the sequence is given by : 3*2(k-1)
a1 = 3 a2 = 6 a4 = 24
So......
3*2(k-1) > 80
2(k-1) > 80/3
k - 1 > log ( 80 / 3) / log (2)
k > log ( 80 / 3) / log (2) + 1
k > 5.737
So k = 6 ...and 3*2(6 - 1) = 96
