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A person is to count 4500 currency notes...

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

Text Solution

Verified by Experts

The correct Answer is:
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.
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