λ1hc=E0z2(nc21−nA21) ....(i) And λ2hc=E0z2(nc21−nB21) So for A and B ....(ii) λ3hc=E0z2(nB21−nA21) Clearly subtracting equation (ii) from equation (i) $\begin{aligned}
& h c\left[\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right]=E_0 z^2\left[\frac{1}{n_B^2}-\frac{1}{n_A^2}\right]=\frac{h c}{\lambda_3} \
& \Rightarrow \frac{1}{\lambda_3}=\frac{1}{\lambda_1}-\frac{1}{\lambda_2} \Rightarrow \frac{1}{\lambda_3}=\frac{(6000-2000)}{6000 \times 2000}=\frac{1}{3000} \
& \lambda_3=3000 Å
\end{aligned}$