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A person is to count 4500 currency notes. Let a n denote the number of notes he counts in the n th minute. If a1 = a 2 = .... = 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

A person is to count 4500 currency notes. Let a n denote the number of notes he counts in the n th minute. If a1 = a 2 = .... = 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

Grade:12

3 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
6 years ago
suppose he takes n minutes to count 4500 notes.
If a1 = a 2 = .... = a 10 = 150
then a1+a2+........+a10=1500
hence a11+a12+.....+an=3000
\frac{n-10}{2}(a11+an)=3000
\frac{n-10}{2}(148+148+(n-11)(-2))=3000
n2-169n+4590=0
n=135,34
so all the notes get counted in 34 minutes
Thanks and Regards
Shaik Aasif
askIITians Faculty
Fkr
13 Points
2 years ago
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Rishi Sharma
askIITians Faculty 646 Points
8 months ago
Dear Student,
Please find below the solution to your problem.

suppose he takes n minutes to count 4500 notes.
If a1 = a 2 = .... = a 10 = 150
then a1+a2+........+a10=1500
hence a11+a12+.....+an=3000
0.5*(n-10)*(a11+an) = 3000
0.5*(n-10)*(148+148+(n-11)(-2)) = 3000
n^2 – 169n + 4590 = 0
n = 135,34
so all the notes get counted in 34 minutes

Thanks and Regards

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