Explanation

(c)
Since, $\frac{2}{a + b},\frac{1}{b},\frac{2}{b + c}$ are in AP.
$\Rightarrow \ \frac{2}{b} = 2\left( \frac{1}{a + b} + \frac{1}{b + c} \right)$
$\Rightarrow \frac{1}{b} = \left( \frac{2b + a + c}{ab + ac + b^{2} + bc} \right)$
$\Rightarrow ab + ac + b^{2} + bc = 2b^{2} + ab + bc$
$\Rightarrow b^{2} = ac$
$\Rightarrow a,b,c$ are in GP.