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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 a_{1}=a_{2}=...=a_{10}=150\; and\; a_{10},a_{11},...are in an A.P. with common difference -2, then the time (in minutes) taken by him to count all notes is

A. 24

B. 34

C. 125

D. Ā135

Answers (1)

The number of notes counted in first 10 minutes =150×10=1500
Suppose, the person counts the remaining 3000 currency notes in n minutes. 

Then, 3000= Sum of n terms of an A.P. with first term 148 and common difference −2
⇒3000=2n?{2×148+(n−1)×(−2)}
⇒3000=n(149−n)
⇒n2−149n+3000=0
⇒(n−125)(n−24)=0
⇒n=125,24
Clearly, n=125 is not possible.
Total, time taken=(10+24)=34 min.

Posted by

Satyajeet Kumar

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