Geometric progressions

3+6+12+......384
Find the sum
I've worked out the ratio is 2, what else do I do

Yes, add them all up now.

3 + 6 + 12 + 24 + 48

is there not a specific formula?

Yes, it's a shortcut. If you can't perform the shortcut add them all up manually.

$S_n=\frac{a(r^n-1)}{r-1}[br][br]a=3[br]r=2$

Find n by substituting 384 into the equation for a geometric series

$a_n=a(r)^{n-1}$

Can you hint more at what to do? I'm so confused

The sum is given by the formula she mentioned. The sum of the first $n$ terms of a geometric sequence with common ratio $r$ and first term $a$ is given by $S_n=\frac{a(1-r^n)}{1-r}$

You found $r=2$ which is correct, and you know that $a=3$

The only thing left to do is find $n$ and to do that you need to use the formula for the $n^{th}$ term of a geometric sequence which is $u_n=ar^{n-1}$. So you would have $384=3\cdot 2^{n-1}$ because 384 is the last term, and solve for $n$ to see up to which term of the sequence you are summing up.

Then just substitute $a, r, n$ into the summation formula.