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 $
$\Rightarrow n^{2} - 169n + 4590 = 0$
$\Rightarrow n^{2} - 135\, n -34n + 4590 = 0$
$\Rightarrow \left(n - 135\right) \left(n - 34\right) = 0$
$\Rightarrow n = 135$, $34$
All the notes get counted in 34 minutes.