A person is to cout 4500 currency notes. Let `a_(n)` denotes 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 AP with common difference `-2`, then the time taken by him to count all notes is

Options

A. 34 min
B. 125 min
C. 135 min
D. 24 min

Explanation

To solve the problem step by step, we will break down the information given and use the properties of arithmetic progressions (AP).

### Step 1: Understand the counting in the first 10 minutes
The person counts 150 notes each minute for the first 10 minutes. Therefore, the total number of notes counted in the first 10 minutes is:

\[ a_1 = a_2 = \ldots = a_{10} = 150 \]

Total notes counted in 10 minutes:

\[ \text{Total notes in 10 minutes} = 10 \times 150 = 1500 \]

### Step 2: Calculate the remaining notes
The total number of currency notes is 4500. After counting 1500 notes in the first 10 minutes, the remaining notes are:

\[ \text{Remaining notes} = 4500 - 1500 = 3000 \]

### Step 3: Define the terms from the 11th minute onwards
From the 11th minute onwards, the number of notes counted per minute forms an arithmetic progression (AP) with the first term \(a_{11} = 150 - 2 = 148\) and a common difference \(d = -2\).

The terms from the 11th minute onwards are:

\[ a_{11} = 148, \quad a_{12} = 146, \quad a_{13} = 144, \quad \ldots \]

### Step 4: Write the formula for the sum of the AP
The sum of the first \(n\) terms of an AP can be calculated using the formula:

\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]

Where:
- \(S_n\) is the sum of the first \(n\) terms,
- \(a\) is the first term,
- \(d\) is the common difference,
- \(n\) is the number of terms.

### Step 5: Set up the equation for the remaining notes
We need to find \(n\) such that the sum of the notes counted from the 11th minute onwards equals the remaining notes (3000):

\[ S_n = \frac{n}{2} \times (2 \cdot 148 + (n-1)(-2)) = 3000 \]

### Step 6: Simplify the equation
Substituting the values into the equation:

\[ \frac{n}{2} \times (296 - 2n + 2) = 3000 \]

This simplifies to:

\[ \frac{n}{2} \times (298 - 2n) = 3000 \]

Multiplying both sides by 2:

\[ n(298 - 2n) = 6000 \]

### Step 7: Rearranging the equation
Rearranging gives us:

\[ 2n^2 - 298n + 6000 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Where \(a = 2\), \(b = -298\), and \(c = 6000\):

\[ b^2 - 4ac = (-298)^2 - 4 \cdot 2 \cdot 6000 \]
\[ = 88804 - 48000 = 40804 \]

Calculating \(n\):

\[ n = \frac{298 \pm \sqrt{40804}}{4} \]
\[ \sqrt{40804} = 202 \]
\[ n = \frac{298 \pm 202}{4} \]

Calculating the two possible values for \(n\):

1. \(n = \frac{500}{4} = 125\)
2. \(n = \frac{96}{4} = 24\)

### Step 9: Determine the valid solution
Since \(n = 125\) would lead to negative notes counted, we discard it. Thus, \(n = 24\).

### Step 10: Calculate the total time taken
The total time taken is the time for the first 10 minutes plus the additional \(n\) minutes:

\[ \text{Total time} = 10 + 24 = 34 \text{ minutes} \]

### Final Answer
The total time taken by the person to count all the notes is **34 minutes**.

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