Calculate the wavelength of the first and the last line in the Balmer series of hydrogen spectrum?

Explanation

For the first line in Balmer series `n_(1)=2` and `n_(2)=3`
`(1)/(lambda_(1))=109,677((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))=109677[(1)/((2)^(2))-(1)/((3)^(2))]`
`=109677xx(5)/(36)cm^(-1)=15232.98cm^(-1)`
`lambda_(1)=6.565xx10^(-5)cm=656.5nm`
For the last line in Balmer series `n_(1) = 2` and `n_(2)=oo`
`(1)/(lambda_(1))=109677((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))=109677[(1)/((2)^(2))-(1)/(oo)]`
`=109677xx(1)/(4)cm^(-1)=27419.3cm^(-1)`
`lambda_(1)=3.647xx10^(-5)cm=364.7nm`