Determination of initial excited state via luminescence
Sometimes formulae are more hindrance than help, and perhaps there are some red herrings in your problem. 




  A one electron species initially in some excited state $n_i$ is
  irradiated with a light of wavelength 121 nm when the electron is
  promoted to a further higher orbit $n_f$. In returning back to the
  ground state, it gives an emission spectrum containing 15 bright
  lines. Out of these 15 lines, 9 lines were found to have wavelengths
  smaller than 121 nm while 5 lines have wavelengths greater than 121
  nm. What is the value of $n_i$?


There being fifteen bright lines suggest that $n_f = 6$, since $15 = 5 + 4 + 3 + 2 + 1$, there being $5$ transitions from the $n = 6$ state to $n = 5,\,4,\,3,\,2,\,1$; $4$ transitions from the $n = 5$ state, and so on. This is the basis for the formula you've employed.

Long wavelengths correspond to low energy; we see that the transition from $n_i$ to $n = 6$ is the sixth lowest in energy. $E \propto n_i^{-2}-n_f^{-2}$, so we pull out our handy-dandy calculator and work out: $$ 6 \to 5: 0.012,\,5 \to 4: 0.023,\,6 \to 4: 0.035,\, 4 \to 3: 0.049,\, 5 \to 3: 0.071,\, 6 \to 3: 0.083.$$ Thus $n_i = 3$. We did not need to use the exact wavelength of light used for irradiation and we did not need to know the identity of the species, because this problem is only concerned with the relative energy differences between energy levels.