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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 |
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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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