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A person is to count 4500 currency notes...

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_(10) = 150` and `a_(11) , a_(12),"……….."` are in an A.P. with common difference -2, then the time taken by him to count at notes is `:`

A

24 mintures

B

34 minutes

C

125 minutes

D

135 minutes

Text Solution

Verified by Experts

The correct Answer is:
B
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Knowledge Check

  • Let a_(n) be the nth term of an A.P. If sum_(r=1)^(100) a_(2r) = alpha and sum_(r = 1)^(100) a_(2r-1) = beta , then the common difference of the A.P. is :

    A
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  • The number of ways in which ten candidates A_(1), A_(2), .. .., A_(10) can he ranked, if A_(1) is always above A_(2) is

    A
    `2 xx 8 !`
    B
    `9 !`
    C
    `10 !`
    D
    `5 xx 9 !`

  • Let n in N . If (1+x)^(n)=a_(0)+a_(1)x+a_(x)x^(2)+ . . . .+a_(n) x^(n) and a_(n-3),a_(n-2),a_(n-2),a_(n-1) are in A.P. then:

    A
    `a_(1),a_(2),a_(3)` are in A.P.
    B
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    C
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    D
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