Let $$3, a, b, c$$ be in A.P. and $$3, a-1, b+1, c+9$$ be in G.P. Then, the arithmetic mean of $$a$$, $$b$$ and $$c$$ is:

Given that 3, a, b, c are in A.P., the common difference is constant. Let the common difference be d. Then:

a = 3 + d

b = 3 + 2d

c = 3 + 3d

Also, 3, a-1, b+1, c+9 are in G.P. Substituting the expressions:

a-1 = (3 + d) - 1 = 2 + d

b+1 = (3 + 2d) + 1 = 4 + 2d

c+9 = (3 + 3d) + 9 = 12 + 3d

The sequence is 3, 2+d, 4+2d, 12+3d. In a G.P., the ratio between consecutive terms is constant. Setting the ratio between the second and first term equal to the ratio between the third and second term:

$$\frac{2+d}{3} = \frac{4+2d}{2+d}$$

$$(d - 4)(d + 2) = 0$$

So, d = 4 or d = -2.

For d = 4:

a = 3 + 4 = 7

b = 3 + 2(4) = 11

c = 3 + 3(4) = 15

The G.P. sequence is 3, 7-1=6, 11+1=12, 15+9=24. Ratios: 6/3=2, 12/6=2, 24/12=2, which is a valid G.P. with common ratio 2.

For d = -2:

a = 3 + (-2) = 1

b = 3 + 2(-2) = -1

c = 3 + 3(-2) = -3

The G.P. sequence is 3, 1-1=0, -1+1=0, -3+9=6. The ratio between second and first term is 0/3=0, but between third and second term is 0/0, which is undefined. Hence, this is invalid.

Thus, only d = 4 is valid, giving a = 7, b = 11, c = 15.

$$\frac{a + b + c}{3} = \frac{7 + 11 + 15}{3} = \frac{33}{3} = 11$$