A man counts 4500 currency notes. Let a_n den | vedclass

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(B) The total number of notes is $4500$. For the first $10$ minutes,he counts $150$ notes per minute,so total notes $= 150 	imes 10 = 1500$. Remainin

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$34$

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(B) The total number of notes is $4500$. For the first $10$ minutes,he counts $150$ notes per minute,so total notes $= 150 	imes 10 = 1500$. Remaining notes $= 4500 - 1500 = 3000$. Let $n$ be the number of additional minutes after the first $10$ minutes. The sequence of notes counted from the $11^{th}$ minute onwards is an arithmetic progression with first term $a = 148$ and common difference $d = -2$. The sum of $n$ terms is given by $S_n = \frac{n}{2} [2a + (n-1)d]$. $3000 = \frac{n}{2} [2(148) + (n-1)(-2)]$. $3000 = \frac{n}{2} [296 - 2n + 2] = \frac{n}{2} [298 - 2n] = n(149 - n)$. $n^2 - 149n + 3000 = 0$. $(n - 24)(n - 125) = 0$. Since $n$ must be small enough for the count to remain positive,$n = 24$. Total time $= 10 + 24 = 34$ minutes.

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