In a geometric sequence, the ratio between two consecutive numbers is same.

ratio r = `(a2)/(a1)`

To find nth term of  the sequence we use an formula. `an = a1 . r^(n-1)`

1. 2500, 500, 100,.....

here r =

We need to find `a4, a5` and `a6` of the sequence

`a4 = a1. r^(4-1) = 2500 . (1/5)^3 = 2500/125 = 20`

`a5 = a1. r^(5-1) = 2500 .(1/5)^4 = 2500/625 = 4`

`a6 = a1. r^(6-1) = 2500 .(1/5)^5 = 2500/3125 = 4/5 = 0.8`

`therefore` the next three terms of the sequence are 20, 4, 4/5

2. 2, 6, 8, .......

here 

`a4 =a1. r^(4-1) = 2 . 3^3 = 2. 27 = 54`

`a5 = a1. r^(5-1) = 2. 3^4 = 2. 81 = 162 `

`a6 = a1. r^(6-1) = 2. 3^5 = 2. 243 = 486`

`therefore` the next three terms of the sequence are 54, 162, 486

In a geometric series, the ratio of any two successive numbers is same and we can use that to determine the numbers in the series.

In case 1, the numbers are 2500, 500 and 100. Let us find the constant ratio between these number. 

The ratio between 2500 and 500 is 1/5 (= 500/2500). Similarly, the ratio between 500 and 100 is also 1/5 ( = 100/500). Thus the next number in the series would be 20 (= 100 x 1/5), 4 (= 20 x 1/5) and 4/5 or 0.8.

Thus the series is 2500, 500, 100, 20, 4, 4/5,.....

Similarly for the second case, the series is 2, 6, 18,.. And the ratio between two successive terms is 3 (= 6/2 = 18/6). The next number is series would be 54 (= 18 x 3), 162 (= 54 x 3) and 486 ( = 162 x 3).

Thus the series is 2, 6, 18, 54, 162, 486,.....

Hope this helps.