If the sum of the first n terms of an arithmetic progression, whose first term is the sum of the first n positive integers and whose common difference is n, is `(8n^(2)-11n-20)`, then the sum of all the possible values of n is

Answer

To solve the problem step by step, we need to find the sum of all possible values of \( n \) given the conditions of the arithmetic progression (AP).

### Step 1: Understanding the Problem
We know that the first term \( a \) of the AP is the sum of the first \( n \) positive integers, and the common difference \( d \) is \( n \). The sum of the first \( n \) terms \( S_n \) of an AP is given by the formula:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]

### Step 2: Finding the First Term \( a \)
The sum of the first \( n \) positive integers is:
\[ a = \frac{n(n+1)}{2} \]

### Step 3: Setting Up the Sum Formula
Substituting \( a \) and \( d \) into the sum formula:
\[ S_n = \frac{n}{2} \left(2 \cdot \frac{n(n+1)}{2} + (n-1)n\right) \]
This simplifies to:
\[ S_n = \frac{n}{2} \left(n(n+1) + (n^2 - n)\right) \]
\[ S_n = \frac{n}{2} \left(n^2 + n + n^2 - n\right) = \frac{n}{2} \left(2n^2\right) = n^3 \]

### Step 4: Setting Up the Equation
We are given that:
\[ S_n = 8n^2 - 11n - 20 \]
Thus, we have:
\[ n^3 = 8n^2 - 11n - 20 \]

### Step 5: Rearranging the Equation
Rearranging gives us a cubic equation:
\[ n^3 - 8n^2 + 11n + 20 = 0 \]

### Step 6: Finding Roots Using the Rational Root Theorem
We can use the Rational Root Theorem to test possible rational roots. Testing \( n = -1 \):
\[ (-1)^3 - 8(-1)^2 + 11(-1) + 20 = -1 - 8 - 11 + 20 = 0 \]
Thus, \( n = -1 \) is a root.

### Step 7: Polynomial Long Division
Now we divide \( n^3 - 8n^2 + 11n + 20 \) by \( n + 1 \) using polynomial long division:
1. Divide \( n^3 \) by \( n \) to get \( n^2 \).
2. Multiply \( n^2 \) by \( n + 1 \) to get \( n^3 + n^2 \).
3. Subtract: \( (-8n^2 - n^2) + 11n + 20 = -9n^2 + 11n + 20 \).
4. Divide \( -9n^2 \) by \( n \) to get \( -9n \).
5. Multiply \( -9n \) by \( n + 1 \) to get \( -9n^2 - 9n \).
6. Subtract: \( (11n + 9n) + 20 = 20n + 20 \).
7. Divide \( 20n \) by \( n \) to get \( 20 \).
8. Multiply \( 20 \) by \( n + 1 \) to get \( 20n + 20 \).
9. Subtract: \( 0 \).

Thus, we have:
\[ n^3 - 8n^2 + 11n + 20 = (n + 1)(n^2 - 9n + 20) \]

### Step 8: Factoring the Quadratic
Now we need to factor \( n^2 - 9n + 20 \):
\[ n^2 - 9n + 20 = (n - 5)(n - 4) \]

### Step 9: Finding All Roots
Thus, the complete factorization is:
\[ (n + 1)(n - 5)(n - 4) = 0 \]
The roots are:
\[ n = -1, n = 5, n = 4 \]

### Step 10: Validating Roots
Since \( n \) must be a positive integer (as it represents the number of terms), we discard \( n = -1 \).

### Step 11: Sum of Valid Roots
The valid roots are \( n = 5 \) and \( n = 4 \). The sum of all possible values of \( n \) is:
\[ 5 + 4 = 9 \]

### Final Answer
Thus, the sum of all possible values of \( n \) is \( \boxed{9} \).