A cricketer has to score 4500 runs. Let `a _(n)` denotes the number of runs he scores in the `n ^(th)` match. If `a _(1)=a_(2)= …. a _(10) =150` and `a _(10) , a _(11), a_(12)….` are in A.P. with common difference `(-2)`. If N be the total number of matches played by him to scoere 4500 runs. Find the sum of the digits of N.

Answer

To solve the problem step by step, we will break down the information given and calculate the required values.

### Step 1: Calculate Runs Scored in the First 10 Matches
The cricketer scores 150 runs in each of the first 10 matches. Therefore, the total runs scored in these matches is:
\[ \text{Total runs in first 10 matches} = 10 \times 150 = 1500 \]

### Step 2: Calculate Remaining Runs to be Scored
The cricketer needs to score a total of 4500 runs. After scoring 1500 runs in the first 10 matches, the remaining runs to be scored is:
\[ \text{Remaining runs} = 4500 - 1500 = 3000 \]

### Step 3: Identify the Sequence for Matches After the 10th Match
From the problem, we know that the runs scored from the 11th match onwards are in an arithmetic progression (A.P.) with the first term \( a_{11} = 150 - 2 = 148 \) and a common difference \( d = -2 \).

### Step 4: Write the Formula for the Sum of an A.P.
The sum \( S_n \) of the first \( n \) terms of an arithmetic progression can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
In our case, we need to find \( n \) such that:
\[ S_n = 3000 \]
where \( a = 148 \) and \( d = -2 \).

### Step 5: Set Up the Equation
Substituting the values into the sum formula gives:
\[ 3000 = \frac{n}{2} \times (2 \times 148 + (n - 1)(-2)) \]
This simplifies to:
\[ 3000 = \frac{n}{2} \times (296 - 2n + 2) \]
\[ 3000 = \frac{n}{2} \times (298 - 2n) \]
Multiplying both sides by 2 to eliminate the fraction:
\[ 6000 = n(298 - 2n) \]
Rearranging gives:
\[ 2n^2 - 298n + 6000 = 0 \]

### Step 6: Solve the Quadratic Equation
We can use the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 2, b = -298, c = 6000 \):
\[ b^2 - 4ac = (-298)^2 - 4 \cdot 2 \cdot 6000 \]
Calculating:
\[ b^2 = 88804, \quad 4ac = 48000 \]
Thus:
\[ b^2 - 4ac = 88804 - 48000 = 40804 \]
Now, taking the square root:
\[ \sqrt{40804} = 202 \]
Now substituting back into the quadratic formula:
\[ n = \frac{298 \pm 202}{4} \]
Calculating the two possible values for \( n \):
1. \( n = \frac{500}{4} = 125 \)
2. \( n = \frac{96}{4} = 24 \)

### Step 7: Determine Validity of \( n \)
Since \( a_{11} = 148 \) and the terms decrease by 2, we need to check which value of \( n \) is valid:
- For \( n = 125 \): The score would eventually become negative, which is not possible.
- For \( n = 24 \): This is valid as it leads to a non-negative score.

### Step 8: Calculate Total Matches Played
The total number of matches played by the cricketer is:
\[ N = 10 + n = 10 + 24 = 34 \]

### Step 9: Find the Sum of the Digits of \( N \)
The sum of the digits of \( N = 34 \) is:
\[ 3 + 4 = 7 \]

### Final Answer
Thus, the sum of the digits of \( N \) is:
\[ \boxed{7} \]