Let the three consecutive terms of the G.P. be $\frac{a}{r},\ a,\ ar$ where $a$ is the middle term.

Use the product condition The product of these three terms is: $\frac{a}{r}\cdot a \cdot ar = a^3$ Given that this product is $512$, we get: $a^3 = 512$ $a = 8$

So the three terms are: $\frac{8}{r},\ 8,\ 8r$

Use the A.P. condition after adding 4 to the first and second terms New terms become: $\frac{8}{r}+4,\ 12,\ 8r$

For these to be in A.P., the middle term is the average of the first and third: $2\cdot 12 = \left(\frac{8}{r}+4\right) + 8r$ $24 = \frac{8}{r} + 8r + 4$ $20 = \frac{8}{r} + 8r$ Divide by $4$: $5 = \frac{2}{r} + 2r$ Multiply by $r$: $5r = 2 + 2r^2$ $2r^2 - 5r + 2 = 0$

Solve the quadratic $2r^2 - 5r + 2 = 0$ $(2r-1)(r-2)=0$ Hence, $r = \frac12 \quad \text{or} \quad r = 2$

Find the original terms and their sum If $r=2$, the G.P. terms are: $4,\ 8,\ 16$ Sum: $4+8+16=28$

If $r=\frac12$, the terms are: $16,\ 8,\ 4$ Sum: $16+8+4=28$

So in either case, the sum of the original three terms is: $\boxed{28}$

Compare with stored answer The stored correct answer is B, i.e. $28$.

Hence, my derived answer agrees with the stored answer.