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 :

<p>Since $3, a, b, c$ are in arithmetic progression (A.P.), the common difference can be calculated using the term $a$ (the second term) as follows:</p> <p>$d = a - 3$</p> <p>The nth term of an A.P. is given by the formula:</p> <p>$T_n = a + (n-1)d$</p> <p>So, using this formula, we can express $b$ and $c$ in terms of $a$ and $d$:</p> <p>$b = a + d$</p> <p>$c = a + 2d$</p> <p>Substituting $d = a - 3$ into these expressions:</p> <p>$b = a + (a - 3)$</p> <p>$c = a + 2(a - 3)$</p> <p>Therefore:</p> <p>$b = 2a - 3$</p> <p>$c = 3a - 6$</p> <p>Now, let's consider that $3, a-1, b+1, c+9$ are in geometric progression (G.P.). For terms in a G.P., the ratio (common ratio, r) between consecutive terms is constant. So:</p> <p>$\frac{a - 1}{3} = \frac{b + 1}{a - 1} = \frac{c + 9}{b + 1}$</p> <p>Now, we will establish the relation between the terms using the property of G.P.:</p> <p>$\frac{a - 1}{3} = \frac{b + 1}{a - 1}$</p> <p>$(a - 1)^2 = 3(b + 1)$</p> <p>$a^2 - 2a + 1 = 3b + 3$</p> <p>Substituting $b = 2a - 3$, we get:</p> <p>$a^2 - 2a + 1 = 3(2a - 3) + 3$</p> <p>$a^2 - 2a + 1 = 6a - 9 + 3$</p> <p>$a^2 - 8a + 7 = 0$</p> <p>Solving this quadratic equation:</p> <p>$(a - 7)(a - 1) = 0$</p> <p>Hence, $a = 7$ or $a = 1$. However, if $a = 1$, the terms $3, a-1, b+1, c+9$ cannot form a G.P. as it would involve division by zero. Therefore, $a = 7$. We use this value to find $b$ and $c$:</p> <p>$b = 2a - 3 = 2(7) - 3 = 14 - 3 = 11$</p> <p>$c = 3a - 6 = 3(7) - 6 = 21 - 6 = 15$</p> <p>Now we can find the arithmetic mean ($A$) of $a$, $b$, and $c$:</p> <p>$A = \frac{a + b + c}{3}$</p> <p>$A = \frac{7 + 11 + 15}{3}$</p> <p>$A = \frac{33}{3}$</p> <p>$A = 11$</p> <p>Hence, the arithmetic mean of $a$, $b$, and $c$ is $11$, which corresponds to Option D.</p>