Question

the fourth and seventh term of a geometric series are 1 / 8

Answer 1

let first term be a and ratio be r

given  fourth and seventh term of a geometric series are 1 / 18 and -1/486

i.e.,ar³=1/18

a=1/(18r³)----->1

(ar³)²=-1/486

a=-1/(486r^6)---------->2  

from 1 and 2 we get -1/(486r^6)=1/(18r³)

from this we get r=-1/3

we know ar³=1/18

=>a(-1/3)³=1/18

from this we get a=-3/2

Hence series is a,ar,ar²,ar³,...........

-3/2,1/2,-1/6,-1/18,..............