A person is to count 4500 currency notes. Let $${a_n}$$ denote the number of notes he counts in the $${n^{th}}$$ minute. If $${a_1} = {a_2} = .... = {a_{10}} = 150{\text{ and }}{a_{10}},{a_{11}},....$$         are in an A.P. with common difference $$- 2$$ , then the time taken by him to count all notes is

Solution :
Till $${10^{th}}$$ minute number of counted notes = 1500
$$\eqalign{ & 3000 = \frac{n}{2}\left[ {2 \times 148 + \left( {n - 1} \right)\left( { - 2} \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\, = n\left[ {148 - n + 1} \right] \cr & {n^2} - 149n + 3000 = 0 \cr & \Rightarrow \,\,n = 125,24 \cr & {\text{But }}n = 125{\text{ is not possible}} \cr & \therefore \,\,{\text{total time = }}24 + 10 = 34{\text{ minutes}}{\text{.}} \cr} $$