3, 6, 12, 25

a1 = 3

a2= 3*2 = 6

a3 = 3*2^2 = 12

a4 = 3*2^3 = 24

We notice that we have a Geometric progression with constant difference r = 2

==> an = 3*r^(n-1)

Now let us calculate 5th, 6th, and 7th terms:

a5 = 3*r^(5-1) = 3*2^4 = 48

a6 = 3*r^(6-1) = 3*2^5 = 96

a7 = 3*r^(7-1) = 3*2^6 = 192

Then sum of ba5, a6, and a7 is:

S = a5 +a6 + a7

= 48 +96 + 192

= 336