A circle with area A ${cm}^{2}$ is contained in the interior of a larger circle with area A + B. Now the radius of the larger circle is 4cm. If A, B, A + B are in arithmetic progression, then the diameter (in cm) of the smaller circle is
$\begin{align}
  & \text{a) }\dfrac{\sqrt{3}}{2} \\
 & \text{a) }\dfrac{\sqrt{3}}{2} \\
 & \text{b) }\dfrac{4\sqrt{3}}{3} \\
 & \text{c) }\dfrac{8\sqrt{3}}{3} \\
 & d)\text{ 2}\sqrt{3} \\
\end{align}$

Hint: Now we know that the area of circle is \[\pi {{r}^{2}}\]. Hence we can find the area of the larger circle and this will be A+B. Hence we get our first equation in A and B. Also, we know that A, B, A+B. are in AP. hence the difference between two successive terms is equal. From this, we will get $2^{nd}$ equation in A and B. Solving the two-equation we will find the value of A.

Complete step-by-step solution:

Now let us consider the bigger circle.
The area of the bigger circle is given by A + B. Now we know that area of the circle is given by $\pi {{r}^{2}}$ where r is the radius of the circle. Since it is given that the radius of the larger circle is 4cm. we get
$\pi {{\left( 4 \right)}^{2}}=A+B$
Hence we have $A+B=16\pi ................(1)$
Now it is given that A, B, and A+B are in arithmetic progression
This means the difference between successive terms is the same
Hence we get $A + B – B = B – A $
Taking A to LHS we get, $A + A = B$
Hence we have 2A = B ………………. (2)
Now from equation (1) and equation (2) we get
$A + 2A = 16\pi $
Hence, $3A=16\pi $, taking 3 on RHS we get.
$A=\dfrac{16}{3}\pi $
Now we know that A is the area of a smaller circle. Let us say r is the radius of the smaller circle.
Now since Area of circle is given by $\pi {{r}^{2}}$ we have
$\pi {{r}^{2}}=\dfrac{16}{3}\pi $
Now dividing the whole equation by π we get
$\begin{align}
  & {{r}^{2}}=\dfrac{16}{3} \\
 & \Rightarrow r=\dfrac{\sqrt{16}}{\sqrt{3}}=\dfrac{4}{\sqrt{3}} \\
\end{align}$
Hence radius of smaller circle is $\dfrac{4}{\sqrt{3}}$
Now diameter of the circle is 2r = $\dfrac{2(4)}{\sqrt{3}}=\dfrac{8}{\sqrt{3}}$
Now multiplying numerator and denominator with $\sqrt{3}$ we get
$\begin{align}
  & =\dfrac{8\sqrt{3}}{\sqrt{3}\left( \sqrt{3} \right)} \\
 & =\dfrac{8\sqrt{3}}{3} \\
\end{align}$
Hence the diameter of the smaller circle is $\dfrac{8\sqrt{3}}{3}$
Option c is the correct option.

Note: Now note the difference in AP is given by $d={{t}_{n}}-{{t}_{n-1}}$ . Hence when a, b, c are in AP then we have $b-a=c-b$. Take care of the order while writing the difference. Also, we can directly take the Arithmetic to mean to form an equation. Hence If A, B, A + B are in the arithmetic progression we can directly write $A+A+B=2B$ and hence we will have the equation B = 2A