Question

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

• A. $34$ minutes
• B. $125$ minutes
• C. $135$ minutes
• D. $24$ minutes

Answer 1

Answer: (A)

$34$ minutes

Till ${10}^{th}$ minutes number of counted notes $=1500$

since, $a_1=a_2=a_3=a_4=a_5=a_6=a_7=a_8=a_9=a_{10}=150$
now, from $a_{10}$ it becomes a A.P. with common diff$=-2$

For, the rest no. of $3000$ notes

$\therefore a_1=a_{11}=148$

$3000=\frac{n}{2}[2\times 148+(n-1\left)\right(-2\left)\right]=n[148-n+1]$

$\Rightarrow n^2-149n+3000=0$

$\Rightarrow n=125,24$

as on putting $n=125$, we get $a_n=148+(125-1)-2=$-ve number, which is not possible

$n=125$ is not possible.

Total time $=24+10=34$ minutes.