Let 3, a, b, c be in A . P . and 3, a - 1, b + 1, c + 9 be in G . P...

Given: $3, a, b, c$ are in $A . P$ 

So, the common difference will be $d = a - 3$

Now, $b = 2a - 3$ and $c = 3a - 6$

Also, $3, a - 1, b + 1, c + 9$ are in $G . P$

So, $3\left(c + 9\right) = \left(b + 1\right)\left(a - 1\right)$

$⇒ 3\left(3a - 6 + 9\right) = \left(2a - 3 + 1\right)\left(a - 1\right)$

$⇒ 3\left(3a + 3\right) = \left(2a - 2\right)\left(a - 1\right)$

$⇒ 9a + 9 = 2{a}^{2} - 4a + 2$

$⇒ 2{a}^{2} - 13a - 7 = 0$

$⇒ a = 7, \dfrac{- 1}{2}\left\{\text{rejected}\right\}$

So, $b = 2 \times 7 - 3 = 11 \text{ and } c = 3 \times 7 - 6 = 15$

Now, the $A . M$ of $a, b, c$ is given by,

$⇒ A = \dfrac{a + b + c}{3}$

$⇒ A = \dfrac{7 + 11 + 15}{3}$

$⇒ A = \dfrac{33}{3}$

$⇒ A = 11$