Let $${a_1},{a_2},{a_3},.....$$    be in harmonic progression with $${a_1} = 5$$   and $${a_{20}} = 25$$  . The least positive integer $$n$$ for which $${a_n} < 0$$   is

Solution :
$$\eqalign{ & \because \,\,{a_1},{a_2},{a_3},.....{\text{ are in H}}{\text{.P}}{\text{.}} \cr & \therefore \,\,\frac{1}{{{a_1}}},\frac{1}{{{a_2}}},\frac{1}{{{a_3}}}.....{\text{ are in A}}{\text{.P}}{\text{.}} \cr & \therefore \,\,\,\,\frac{1}{{{a_1}}} = \frac{1}{5}{\text{ and }}\frac{1}{{{a_{20}}}} = \frac{1}{{25}} \cr & \frac{1}{{{a_1}}} + 19d = \frac{1}{{{a_{20}}}} \cr & \Rightarrow \,\,\frac{1}{5} + 19d = \frac{1}{{25}} \cr & \Rightarrow \,\,d = \frac{{ - 4}}{{475}} \cr & {\text{Now }}\frac{1}{{{a_n}}} = \frac{1}{5} + \left( {n - 1} \right)\left( {\frac{{ - 4}}{{475}}} \right) \cr & {\text{Clearly }}\,{a_n} < 0{\text{ }}\,{\text{if }}\,\frac{1}{{{a_n}}} < 0 \cr & \Rightarrow \,\,\frac{1}{5} - \frac{{4n}}{{475}} + \frac{4}{{475}} < 0 \cr & \Rightarrow \,\, - 4n < - 99{\text{ or }}n > \frac{{99}}{4} = 24\frac{3}{4} \cr & \therefore \,\,n \geqslant 25 \cr & {\text{Hence least value of }}n{\text{ is 25}}{\text{.}} \cr} $$