Let `a _(1), a_(2), a_(3),…….., a_(n)` be real numbers in arithmatic progression such that `a _(1) =15 and a_(2)` is an integer. Given ` sum _( r=1) ^(10) (a_(r)) ^(2) =1185`. If `S_(n) = sum _(r =1) ^(n) a_(r) and ` maximum value of `n` is `N` for which `S_(n) ge S_((n +1)), ` then find `N -10.`

Answer

To solve the problem step by step, we will break down the information given and derive the necessary equations.

### Step 1: Understand the Arithmetic Progression
We are given that \( a_1 = 15 \) and \( a_2 \) is an integer. The general term of an arithmetic progression (AP) can be expressed as:
\[ a_r = a_1 + (r-1)d \]
where \( d \) is the common difference.

### Step 2: Write the Expression for \( S_n \)
The sum of the first \( n \) terms of the AP is given by:
\[ S_n = \sum_{r=1}^{n} a_r = n \cdot a_1 + \frac{(n-1)n}{2} d \]

### Step 3: Calculate \( \sum_{r=1}^{10} a_r^2 \)
We know:
\[ \sum_{r=1}^{10} a_r^2 = 1185 \]
Substituting the expression for \( a_r \):
\[ \sum_{r=1}^{10} (a_1 + (r-1)d)^2 = \sum_{r=1}^{10} (15 + (r-1)d)^2 \]
Expanding this:
\[ = \sum_{r=1}^{10} (15^2 + 2 \cdot 15 \cdot (r-1)d + (r-1)^2d^2) \]
\[ = 10 \cdot 225 + 2 \cdot 15d \sum_{r=1}^{10} (r-1) + d^2 \sum_{r=1}^{10} (r-1)^2 \]

### Step 4: Calculate the Summations
1. \( \sum_{r=1}^{10} (r-1) = 0 + 1 + 2 + \ldots + 9 = \frac{9 \cdot 10}{2} = 45 \)
2. \( \sum_{r=1}^{10} (r-1)^2 = 0^2 + 1^2 + 2^2 + \ldots + 9^2 = \frac{9 \cdot 10 \cdot 19}{6} = 285 \)

### Step 5: Substitute Back into the Equation
Substituting these values back:
\[ 1185 = 2250 + 2 \cdot 15d \cdot 45 + 285d^2 \]
\[ 1185 = 2250 + 1350d + 285d^2 \]
Rearranging gives:
\[ 285d^2 + 1350d + 2250 - 1185 = 0 \]
\[ 285d^2 + 1350d + 1065 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ d = \frac{-1350 \pm \sqrt{1350^2 - 4 \cdot 285 \cdot 1065}}{2 \cdot 285} \]
Calculating the discriminant:
\[ 1350^2 = 1822500 \]
\[ 4 \cdot 285 \cdot 1065 = 1214100 \]
\[ \sqrt{1822500 - 1214100} = \sqrt{608400} = 780 \]
Thus:
\[ d = \frac{-1350 \pm 780}{570} \]
Calculating the two possible values for \( d \):
1. \( d = \frac{-570}{570} = -1 \)
2. \( d = \frac{-2130}{570} = -3.7368 \) (not an integer)

### Step 7: Determine \( N \)
Since \( d = -1 \), we have:
\[ a_n = 15 + (n-1)(-1) = 16 - n \]
To find \( N \) such that \( S_n \geq S_{n+1} \):
\[ S_{n+1} - S_n = a_{n+1} = 16 - (n + 1) < 0 \]
This gives:
\[ 16 - (n + 1) < 0 \implies n > 15 \]
The maximum integer \( n \) is 16.

### Step 8: Calculate \( N - 10 \)
Thus:
\[ N - 10 = 16 - 10 = 6 \]

### Final Answer
The final answer is:
\[ \boxed{6} \]