1.

A person is to cout 4500 currency notes. Let `a_(n)` denotes 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 AP with common difference `-2`, then the time taken by him to count all notes isA. 34 minB. 125 minC. 135 minD. 24 min

Answer» Correct Answer - A
Till 10th minute, number of counted notes =1500
`:. 3000=(n)/(2){2xx148+(n-1)xx-2}=n(148-n+1)`
`implies n^(2)-149n+3000=0`
`implies (n-125)(n-24)=0`
`:.n=125,24`
`n=125` is not possible.
`:.` n=24`
`:. " Total time =10+24=34 min`.


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