A person is to count 4500 currency notes let an denotes the numbers of notes he counts in the nth minute. If ${\mathrm{a}}_1={\mathrm{a}}_2=\dots \dots {\mathrm{a}}_{10}=150\text{ and }{\mathrm{a}}_{10},{\mathrm{a}}_{11}\dots \dots$are in A.P with Common difference -2 then time taken by him to count all notes is

${\mathrm{a}}_1={\mathrm{a}}_2={\mathrm{a}}_3=\dots \dots {\mathrm{a}}_{10}=150$
 No of notes counted till 10th minute is
$$\begin{gathered} \begin{array}{l}=10\times 150=1500 \\ \text{Remaining notes }=3000 \\ \mathrm{Sn}=3000\Rightarrow \frac{\mathrm{n}}{2}[2.148+(\mathrm{n}-1\left)\right(-2\left)\right]=3000\\ \Rightarrow 148\mathrm{n}-\mathrm{n}(\mathrm{n}-1)=3000 \\ \Rightarrow {\mathrm{n}}^2-\mathrm{n}-148\mathrm{n}+3000=0 \\ \Rightarrow (\mathrm{n}-125)(\mathrm{n}-24)=0\\ \Rightarrow \mathrm{n}=24\text{ (or) }125\text{ (rejected) } \\ \therefore \mathrm{n}=24 \\ \text{Total time is }24+10=34\min \end{array} \end{gathered}$$