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

Options

A. 24 minutes
B. 34 minutes
C. 125 minutes
D. 135 minutes

Answer

B

Explanation

Given, `a_(1)=a_(2)=a_(3)=......=a_(10)=150`
Also, `a_(10),a_(11),a_(12),...." are in A.P. and "D=-2`
Since, `a_(10)=150,A.P." is "150,148,146,....`
For the first 10 minutes, he has counted `150xx10=1500` notes.
Time taken to count remaining 3000 notes
`S_(n)=(n)/(2)[2a+(n-1)d]`
`implies3000=(n)/(2)[2xx148+(n-1)(-2)]`
`implies3000=(n)/(2)xx2(148-n+1)`
`implies3000=148n-n^(2)+n`
`impliesn^(2)-149n+3000=0`
`implies(n-24)(n-125)`
`impliesn=24," or "n=125.`
Since he has taken 10 minutes to count 1500 notes, he will not take 125 min to count 3000 notes.
So, `n=24.`
`:." Total time"=10+24=34" minutes."`