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A person is to count 4500 currency notes...

A person is to count 4500 currency notes. Let an 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 A.P. with common difference -2, then the time taken by him to count all notes is (1) 34 minutes (2) 125 minutes (3) 135 minutes (4) 24 minutes

A

125 minutes

B

135 minutes

C

24 mintutes

D

34 minutes

Text Solution

Verified by Experts

The correct Answer is:
D
The number of notes counted in first 10 minutes `=150xx10=1500`
Suppose the person counts the remaining 3000 currency notes in n minutes. Then,
3000=Sum of terms of an AP with first term 148 and common difference -2
`rArr" "3000=(n)/(2){2xx148+(n-1)xx(-2)}`
`rArr" "3000=n(149-n)`
`rArr" "n^(2)-149n+3000=0`
`rArr" "(n-125)(n-24)=0rArrn=125,24`
Clearly, n-125 is not possible.
`:.` Total time taken =(10+42)mintutes =34 minutes.
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