The emission spectrum of hydrogen atoms has two lines from the Balmer series with wavelengths 4102 Å and 4861 Å. The wave number of another emission line is equal to the difference between the wave numbers of these two lines. Find the $\lambda$ of  line and series to which it belongs

Wave number = $\frac{1}{\lambda }={\mathrm{RZ}}^2\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right)$

where R = Rydberg constant = $1.097\times {10}^7/\mathrm{m}$

Here , $Z=1$, 

Hence   ${\mathrm{RZ}}^2=1.097\times {10}^7/\mathrm{m}$

$\Rightarrow \frac{{10}^{10}}{4102}=1.097\times {10}^7\left[\frac{1}{2^2}-\frac{1}{{\mathrm{n}}_2^2}\right]$

$\text{ (or) }\frac{1}{4}-\frac{1}{{\mathrm{n}}_2^2}=\frac{1000}{4102\times 1.097}=0.2222\text{ (or) }{\mathrm{n}}_2=6$

Similarly   $\frac{{10}^{10}}{4861}=1.097\times {10}^7\left[\frac{1}{2^2}-\frac{1}{{\mathrm{n}}_2^2}\right]\text{ (or) }{\mathrm{n}}_2=4$

Now the wave number of third line = difference in wave number of above two lines 

$\frac{1}{\lambda }={\mathrm{RZ}}^2\left(\frac{1}{4^2}-\frac{1}{6^2}\right)1.097\times {10}^7\times \frac{20}{576}/\mathrm{m}$  

$\lambda =\frac{576}{20}\times \frac{1}{1.097\times {10}^7}=\frac{57600}{2\times 1.097}\mathrm{A}=26253\text{Å}$

Since n₁ is 4 here, the series will be Brackett.