Question

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

• A. 125 minutes
• B. 135 minutes
• C. 34 minutes
• D. 24 minutes

Answer 1

Answer: (C)

Suppose he takes n minutes to count $4500$ notes. We have $a_1+a_2+.....a_{10}=10(150)=1500$ $a_{11}=148$ and $a_{11}$, $a_{12}.........a_n$ is an A.P. with common difference $d=-2$.
We are given
$a_1+a_2+.....+a_{10}+a_{11}+.....+a_n=4500$
$\Rightarrow a_{11}+a_{12}+.......+a_n=3000$.
$\Rightarrow \frac{n-10}{2}\left[a_{11}+a_n\right]=3000$
$\Rightarrow \frac{1}{2}\left(n-10\right)\left[148+148+\left(n-11\right)\left(-2\right)\right]=3000$
$\Rightarrow \left(n-10\right)\left(148-n+11\right)=3000$
$\Rightarrow \left(n-10\right)\left(159-n\right)=3000$
$\Rightarrow n^2-169n+4590=0$
$\Rightarrow n^2-135n-34n+4590=0$
$\Rightarrow \left(n-135\right)\left(n-34\right)=0$
$\Rightarrow n=135$, $34$
All the notes get counted in 34 minutes.