A person is to count 4500 currency notes.

Question

A person is to count 4500 currency notes. Let $a_n$ denote the number of notes he counts in the $n$-th minute. If $a_1=a_2=\cdots=a_{10}=150$ and $a_{10},a_{11},\ldots$ are in an A.P. with common difference $-2$, then the time taken by him to count all the notes is:

Accepted Answer

Notes counted in the first 10 minutes $=1500$; remaining $=3000$.

From minute 11 onward, the count follows an A.P. with first term $148$ (i.e. $a_{10}-2$) and common difference $-2$. For $n$ further minutes:

$$\frac n2\big[2(148)+(n-1)(-2)\big]=3000 \Rightarrow n(149-n)=3000 \Rightarrow n^2-149n+3000=0.$$

$$n=\frac{149\pm\sqrt{149^2-12000}}{2}=\frac{149\pm101}{2}=125\text{ or }24.$$

$n=125$ is rejected (the counting rate would go negative long before then); $n=24$ is valid.

$$\text{Total time}=10+24=34\text{ minutes.}$$

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