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_3=..........=a_10=150` and `a_10,a_11,.........`are in an `AP` with common difference `-2`, then the time taken by him to count all notes is :- (1) 24 minutes 10 11 (2) 34 minutes (3) 125 minutes (4) 135 minutes
A. 135 min
B. 24 min
C. 34 min
D. 125 min

Answer 1

Correct Answer - C
Till `10^(th)` minute, the number of counted notes is 1500.
`therefore3000=n/2[2xx148+(n-1)(-2)]`
=n[148-n+1]
`rArrn^(2)-149n+3000=0`
`thereforen=125,24`
Since n=125 is not possible, total time required is 24+10=34 min.