The angles A, B, C of a ∆ABC are in AP and the sides a, b,

Since A, B, C are A.P

$\Rightarrow 2 B=A+C$

Since $A+B+C=\pi$  (By angle sum property)

$\Rightarrow 3 B=\pi$

i. e $B=\frac{\pi}{3}$     ....(1)

Also,

Since a, b, c are in g.p 

$\Rightarrow b^{2}=a c$   ...(2)

Using $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}$

i.e $\cos \frac{\pi}{3}=\frac{a^{2}+c^{2}-a c}{2 a c} \quad$ from (1) and (2).

$\Rightarrow \frac{1}{2}=\frac{a^{2}+c^{2}-a c}{2 a c}$

$\Rightarrow a c=a^{2}+c^{2}-a c$

$\Rightarrow a^{2}+c^{2}=2 a c$

$\Rightarrow a^{2}+c^{2}=2 b^{2}$   from (2)

Hence $\lambda=2$