Solution:

For Lyman series

$$\upsilon = Rc\left[\frac{1}{1^{2}} -\frac{1}{n^{2}}\right]$$

where $$n = 2, 3,4,......$$

For the series limit of Lyman series, $$n = \infty$$

$$\therefore\upsilon_{1} = Rc\left[\frac{1}{1^{2}} -\frac{1}{\infty^{2}}\right] = Rc\quad...\left(i\right) $$

For the first line of Lyman series, $$n = 2$$

$$\therefore\upsilon_{2} = Rc\left[\frac{1}{1^{2}} -\frac{1}{2^{2}}\right] = \frac{3}{4}Rc\quad...\left(ii\right) $$

For Balmer series

$$\upsilon = Rc\left[\frac{1}{2^{2}} -\frac{1}{n^{2}}\right]$$

where $$n = 3,4, 5....$$

For the series limit of Balmer series, $$n =\infty$$

$$\therefore \upsilon_{3} = Rc\left[\frac{1}{2^{2}} -\frac{1}{\infty^{2}}\right] = \frac{Rc}{4}\quad...\left(iii\right)$$

From equations $$(i), (ii)$$ and $$(iii)$$, we get

$$\upsilon_{1} = \upsilon_{2} +\upsilon_{3} $$ or

$$ \upsilon_{1} -\upsilon_{2} = \upsilon_{3}$$