Home
Class 12
MATHS
A person is to count 4500 currency notes...

A person is to count 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 is

A

34 min

B

125 min

C

135 min

D

24 min

Text Solution

Verified by Experts

The correct Answer is:
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`.
Promotional Banner

Similar Questions

Explore conceptually related problems

If a_(1),a_(2),a_(3),a_(4) and a_(5) are in AP with common difference ne 0, find the value of sum_(i=1)^(5)a_(i) " when " a_(3)=2 .

Let a_(n) be the nth term of an AP, if sum_(r=1)^(100)a_(2r)=alpha " and "sum_(r=1)^(100)a_(2r-1)=beta , then the common difference of the AP is

The number of functions from f:{a_(1),a_(2),...,a_(10)} rarr {b_(1),b_(2),...,b_(5)} is

Let a_n be the n^(t h) term of an A.P. If sum_(r=1)^(100)a_(2r)=alpha & sum_(r=1)^(100)a_(2r-1)=beta, then the common difference of the A.P. is -

Write the first five terms of the sequences in obtain the corresponding series: a_(1)= a_(2)= 1, a_(n)= a_(n-1) + a_(n-2), n gt 2

The number of ways in which 10 condidates A_(1),A_(2),......,A_(10) can be ranked so that A_(1) is always above A_(2) , is

Let a_(1),a_(2),"...." be in AP and q_(1),q_(2),"...." be in GP. If a_(1)=q_(1)=2 " and "a_(10)=q_(10)=3 , then

Find the indicated terms in each of the sequences whose n^(th) term is given: a_(n)= 2n^(2) -1, a_(7)

If a_(1),a_(2),a_(3),"......" be in harmonic progression with a_(1)=5 and a_(20)=25 . The least positive integer n for which a_(n)lt0 is

Find the indicated terms in each of the sequences whose n^(th) term is given: a_(n)= (n-1) (n+2)(n-3), a_(10)