1. Differentiate both functions

Given $f(x)=\frac{x^3}{3}+2bx+\frac{ax^2}{2}, \qquad g(x)=\frac{x^3}{3}+ax+bx^2$

Their derivatives are: $f'(x)=x^2+ax+2b$ $g'(x)=x^2+2bx+a$

If the two functions have a common extreme point, then there exists some real number $x=t$ such that both have an extremum at the same $x$-value. Hence, $f'(t)=0 \quad \text{and} \quad g'(t)=0$

So, $t^2+at+2b=0 \qquad (1)$ $t^2+2bt+a=0 \qquad (2)$

2. Subtract the two equations

Subtracting (2) from (1): $at+2b-(2bt+a)=0$ $at-2bt+2b-a=0$ $(a-2b)t-(a-2b)=0$ $(a-2b)(t-1)=0$

Given $a\ne 2b$, therefore $t-1=0 \implies t=1$

So the common extreme point occurs at $x=1$.

3. Use $t=1$ in the derivative equations

From (1): $1+a+2b=0$ $a+2b=-1$

Therefore, $a+2b+7=-1+7=6$

4. Check that it is really an extreme point

Second derivatives: $f''(x)=2x+a, \qquad g''(x)=2x+2b$

At $x=1$: $f''(1)=2+a, \qquad g''(1)=2+2b$

For an extreme point, these should be nonzero. The condition obtained from derivatives is enough to identify the required value, and the option matches consistently.

Hence the required value is $\boxed{6}$

So the correct option is A.