Question

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

Answer 1

Correct Answer - 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.