I am given a series 3,6,12,24.... What is the sum of the 5th, 6th and 7th terms of this series.

Terms given are : a1 = 3,

a2
= 6.

a3 =12,

a4
=24.

Clearly a2/a1 = a3/a2 = a4/a3 =
2.

Therefore this is a geometric series (GS) with common
ratio r =2

The n th term an of the GS =
a1*r^(n-1).

So a5 = a1*r^(5-1) = 3*2^(5-1) =
48

a6 = a1*r^(6-1) =3*2^(6-1) = 
96

a7 = a1*r^(7-1) = a6 *r = 96*2 =
192.

If the sum Sn up to the nth term is given
by:

Sn = a(r^(n+1)-1)/(r-1). = 3{2^(n) - 1}/(2-1) =
2(2^n)-1).

So S5  = 3[2^5 -1] = 
93

S6 = 3(2^6-1) = 189

S7 =
3(2^7-1) = 381

This could be cross checked from the series
whose actual terms are as below:

So in the Given
GS: 

 3,6,12,24, 48,96,192 ,
the bold types are 5th 6th and 7th terms.