A person is to count 45000 currency notes. Let $a_{n}$ denotes the number of notes he counts in the $n$ th minute. If $a_{1} = a_{2} = ... = 10_{10} = 150$ and $a_{10},a_{11},\ldots$ are in AP with common difference -2, then the time taken by him to count all notes, is

Question

Dr. Abubacker · Accepted Answer

(b)
Number of notes that the person counts in 10 min
$= 10 	imes 150 = 1500$
Since $a_{10},a_{11},a_{12},\ldots$ are in AP with common difference -2
Let $n$ be the time taken to count remaining 3000 notes, then
$\frac{n}{2}\left\lbrack 2 	imes 148 + (n - 1) 	imes - 2 ightbrack = 3000$
$\Rightarrow$ $n^{2} - 149n + 3000 = 0$
$\Rightarrow$ $(n - 24)(n - 125) = 0$
$\Rightarrow$ $n = 24,\ 125$
Then, the total time taken by the person to count all notes.
$= 10 + 24 = 34$ min

Q. A person is to count 45000 currency notes. Let $a_{n}$ denotes the number of notes he counts in the $n$ th minute. If $a_{1} = a_{2} = ... = 10_{10} = 150$ and $a_{10},a_{11},\ldots$ are in AP with common difference -2, then the time taken by him to count all notes, is

Explanation

Reviewed by Dr. Abubacker

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