A person is to count 4500 notes. Let `a_(n)` denote the number of notes the counts in the nth minute. If `a_(1)=a_(2)=a_(3)="_"=a_(10)=150`, and `a_(10),a_(11),a_(12), . . .` are in AP with the common difference -2, then the time taken by him to count all the notes is:

Options

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

Answer

To solve the problem step by step, we follow these calculations:

### Step 1: Calculate the number of notes counted in the first 10 minutes
Given that the person counts 150 notes per minute for the first 10 minutes:
\[ \text{Total notes in first 10 minutes} = 150 \times 10 = 1500 \]

### Step 2: Calculate the remaining notes to be counted
The total number of notes to be counted is 4500. Therefore, the remaining notes after the first 10 minutes is:
\[ \text{Remaining notes} = 4500 - 1500 = 3000 \]

### Step 3: Define the arithmetic progression (AP) for the notes counted after the 10th minute
From the problem, we know:
- \( a_{10} = 150 \)
- \( a_{11} = a_{10} - 2 = 148 \)
- \( a_{12} = a_{11} - 2 = 146 \)
- The common difference \( d = -2 \)

Thus, the sequence of notes counted from the 11th minute onwards is:
\[ a_{11}, a_{12}, a_{13}, \ldots = 148, 146, 144, \ldots \]

### Step 4: Use the formula for the sum of an arithmetic series to find how many additional minutes are needed to count the remaining notes
Let \( n \) be the number of minutes after the 10th minute. The sum of the first \( n \) terms of an AP is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
Where:
- \( a = 148 \) (the first term after the 10th minute)
- \( d = -2 \) (the common difference)

Setting \( S_n \) equal to the remaining notes:
\[ 3000 = \frac{n}{2} \times (2 \times 148 + (n-1)(-2)) \]

### Step 5: Simplify the equation
\[ 3000 = \frac{n}{2} \times (296 - 2n + 2) \]
\[ 3000 = \frac{n}{2} \times (298 - 2n) \]
\[ 6000 = n(298 - 2n) \]
\[ 6000 = 298n - 2n^2 \]
Rearranging gives:
\[ 2n^2 - 298n + 6000 = 0 \]

### Step 6: Solve the quadratic equation using the quadratic formula
The quadratic formula is:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 2, b = -298, c = 6000 \):
\[ n = \frac{298 \pm \sqrt{(-298)^2 - 4 \times 2 \times 6000}}{2 \times 2} \]
Calculating the discriminant:
\[ (-298)^2 = 88804 \]
\[ 4 \times 2 \times 6000 = 48000 \]
\[ \sqrt{88804 - 48000} = \sqrt{40804} = 202 \]
Now substituting back:
\[ n = \frac{298 \pm 202}{4} \]
Calculating the two possible values for \( n \):
1. \( n = \frac{500}{4} = 125 \)
2. \( n = \frac{96}{4} = 24 \)

### Step 7: Determine the valid solution for \( n \)
Since \( n = 125 \) would lead to negative terms in the sequence (as \( a_{10 + 125} \) would be negative), we reject it. Therefore, we take \( n = 24 \).

### Step 8: Calculate the total time taken
The total time taken to count all the notes is:
\[ \text{Total time} = 10 + n = 10 + 24 = 34 \text{ minutes} \]

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

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