If the first term, common ratio and the last term of a G.P. are a, r and l respectively, then the nth term from the end of the G.P. is given by :

1. lr^{n - 1}

2. lr^{1 - n}

3. $\dfrac{l}{r^{1 - n}}$

4. $\dfrac{l}{r^{n + 1}}$

When the G.P. is reversed, the first term becomes l and the new common ratio is the reciprocal of the original common ratio i.e. $\dfrac{1}{r}$.

$\therefore T_n = l \times \Big(\dfrac{1}{r}\Big)^{n - 1} \\[1em] = \dfrac{l}{r^{n - 1}} \\[1em] = lr^{-(n - 1)} \\[1em] = lr^{1 - n}.$

Hence, option 2 is the correct option.