Question

A person is to count $4500$ currency notes. Let an denote the number of notes he counts in the nth minute. If $a_1 = a_2 = ...... = a_{10} = 150$ and $a_{10}, a_{11}, ......$ 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

Answer 2

Till $10^{th}$ minute number of counted notes $= 1500$ $3000 = \frac{n}{2} \left[2 \times 148 + \left(n - 1\right)\left(-2\right)\right] = n\left[148 - n + 1\right]$ $n^{2} - 149n + 3000 = 0$ $n = 125, 24$ $n = 125$ is not possible. Total time $= 24 + 10 = 34$ minutes.