A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth min

Q: Wait, why is the first term of the AP $148$ and not $150$?

That is a brilliant observation! The problem states that for the first $10$ minutes, the rate is constant at $150$. The AP behavior begins at the $11$th minute. Since the common difference is $-2$, the speed at the $11$th minute is the speed at the $10$th minute minus $2$. So, $a_{11} = 150 - 2 = 148$. It is a transition point, not a continuation of the same constant rate.

Q: Why do we reject the root $n=125$?

Mathematically, the quadratic equation $n^2 - 149n + 3000 = 0$ yields two roots: $24$ and $125$. However, physics imposes constraints that algebra sometimes ignores. If you calculate the speed at $n=125$ using the AP formula $a_n = a + (n-1)d$, you get $148 + (124)(-2) = 148 - 248 = -100$. You cannot count negative notes! Always check if your mathematical solution respects the physical reality of the problem.

Q: Can we solve this using the sum of the first $N$ terms of the whole sequence?

You could, but it would be significantly more complex. You would have to define the sequence piecewise and sum the first $10$ terms separately from the rest. It is much cleaner to treat the problem as a sum of two distinct parts: a rectangle of constant work and an AP of decreasing work. Always look for the path of least resistance in your calculations!

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 $a_1 = a_2 = \dots = a_{10} = 150$ and $a_{10}, a_{11}, \dots$ are in an AP with common difference $-2$, then the time taken by him to count all notes is

Options:

(A) 34 minutes

(B) 125 minutes

(C) 135 minutes

(D) 24 minutes

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a massive, daunting stack of $4500$ currency notes. Your task is to count them all, but your performance is governed by two distinct phases of human behavior.

### Phase 1: The Machine Phase

For the first $10$ minutes, you are unstoppable. The problem states that $a_1 = a_2 = \dots = a_{10} = 150$.

You are counting $150$ notes every single minute. The work done here is simply the area of a rectangle:
$$10 	imes 150 = 1500 	ext{ notes}$$

You have cleared a third of your mountain, but the real challenge is yet to come.

### Phase 2: The Fatigue Phase

From the $11$th minute onwards, your speed begins to decay following an Arithmetic Progression (AP). The first term of this new phase is $a_{11} = 150 - 2 = 148$, and your common difference is $d = -2$.

We have $3000$ notes remaining ($4500 - 1500 = 3000$). We need to find the time $n$ it takes to clear these remaining notes using the sum formula for an AP:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$

Substituting our values, we get:
$$3000 = \frac{n}{2}[2(148) + (n-1)(-2)]$$

### The Quadratic Dance

Let us simplify the expression inside the brackets. We have $2(148) = 296$ and $(n-1)(-2) = -2n + 2$. Combining these, we get $298 - 2n$.

Our equation becomes:
$$3000 = \frac{n}{2}[298 - 2n]$$

The $2$ in the denominator cancels perfectly with the $2$ inside the bracket, leading to:
$$3000 = n(149 - n)$$

Expanding this, we arrive at the quadratic equation:
$$n^2 - 149n + 3000 = 0$$

We need two numbers that multiply to $3000$ and add to $-149$. Those numbers are $-24$ and $-125$. Thus, the equation factors as:
$$(n - 24)(n - 125) = 0$$

### The Physical Reality Check

We have two roots: $n = 24$ and $n = 125$. We must determine which one is physically valid.

If we choose $n = 125$, we calculate the speed at that minute:
$$a_{125} = 148 + (124)(-2) = -100$$

A negative counting rate is physically impossible. Therefore, we reject $n = 125$ and accept $n = 24$.

The second phase takes exactly $24$ minutes. Adding the initial $10$ minutes, the total time is:
$$10 + 24 = \mathbf{34 	ext{ minutes}}$$

You have conquered the mountain of notes. Remember, math is the language of reality; always ensure your equations speak the truth of the physical world.

Select Answer:

Visualized Solution

The Counting Challenge

Phase 1: Constant Counting Rate

Notes Counted in Phase 1

Remaining Notes for Phase 2

Phase 2: Decreasing Rate (AP)

Setting up the Sum Equation

Substituting the Values

Simplifying the Equation

Forming the Quadratic Equation

Solving the Quadratic Equation

Selecting the Valid Root

Final Calculation: Total Time

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

Phase 1

Phase 2

The Quadratic Dance

The Physical Reality Check

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