A person is to count `4500` currency notes. Let `a_n`, denote the number of notes he counts in the `nth` minute if `a_1=a_2=a_3=..........=a_10=150` and `a_10,a_11,.........`are in an `AP` with common difference `-2`, then the time taken by him to count all notes is :- (1) 24 minutes 10 11 (2) 34 minutes (3) 125 minutes (4) 135 minutes

Options

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

Explanation

To solve the problem, we need to determine the total time taken by a person to count 4500 currency notes, given the counting pattern described.

### Step-by-Step Solution:

1. **Count Notes in the First 10 Minutes**:
- The person counts 150 notes per minute for the first 10 minutes.
- Total notes counted in the first 10 minutes:
\[ a_1 + a_2 + a_3 + \ldots + a_{10} = 150 \times 10 = 1500 \text{ notes} \]

2. **Calculate Remaining Notes**:
- Total notes to count = 4500 notes.
- Remaining notes after 10 minutes:
\[ 4500 - 1500 = 3000 \text{ notes} \]

3. **Understanding the Arithmetic Progression (AP)**:
- From the 11th minute onwards, the notes counted are in an AP with the first term \( a_{11} \) and a common difference of -2.
- We know \( a_{10} = 150 \), so:
\[ a_{11} = a_{10} - 2 = 150 - 2 = 148 \]
- The sequence of notes counted from the 11th minute onwards will be:
\[ a_{11} = 148, a_{12} = 146, a_{13} = 144, \ldots \]

4. **Finding the Number of Minutes \( n \)**:
- The \( n \)-th term of the AP can be expressed as:
\[ a_n = a_{11} + (n-1)(-2) = 148 - 2(n-1) = 150 - 2n \]
- The total number of notes counted from the 11th minute to the \( n \)-th minute can be calculated using the formula for the sum of an AP:
\[ S_n = \frac{n}{2} \times (a_{11} + a_n) \]
- Here, \( a_n = 150 - 2n \), so:
\[ S_n = \frac{n}{2} \times (148 + (150 - 2n)) = \frac{n}{2} \times (298 - 2n) = 149n - n^2 \]
- We need this sum to equal the remaining notes (3000):
\[ 149n - n^2 = 3000 \]
- Rearranging gives us:
\[ n^2 - 149n + 3000 = 0 \]

5. **Solving the Quadratic Equation**:
- We can use the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 1, b = -149, c = 3000 \):
\[ n = \frac{149 \pm \sqrt{149^2 - 4 \times 1 \times 3000}}{2 \times 1} \]
- Calculate \( 149^2 = 22201 \) and \( 4 \times 3000 = 12000 \):
\[ n = \frac{149 \pm \sqrt{22201 - 12000}}{2} = \frac{149 \pm \sqrt{10201}}{2} = \frac{149 \pm 101}{2} \]
- This gives us two potential solutions:
\[ n = \frac{250}{2} = 125 \quad \text{and} \quad n = \frac{48}{2} = 24 \]

6. **Determining Valid \( n \)**:
- Since \( n = 125 \) is not feasible in the context of the problem (as it would take too long), we take \( n = 24 \).

7. **Total Time Calculation**:
- Total time taken is the time for the first 10 minutes plus the time for the remaining notes:
\[ \text{Total time} = 10 + 24 = 34 \text{ minutes} \]

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