Let a1,a2,a3 ...... a11 be real numbers satisfying `a_1 =15, 27-2a_2 > 0 and a_k= 2a_(k-1) - a_(k-2)` for `k=3,4,.....11` If `(a1^2 +a2^2.......a11^2)/11 = 90` then find the value of `(a_1+a_2....+a_11)/11`

Explanation

To solve the problem step by step, we will analyze the given conditions and derive the necessary values.

### Step 1: Understand the given conditions
We have:
- \( a_1 = 15 \)
- \( 27 - 2a_2 > 0 \) implies \( a_2 < \frac{27}{2} = 13.5 \)
- The recurrence relation \( a_k = 2a_{k-1} - a_{k-2} \) for \( k = 3, 4, \ldots, 11 \)

### Step 2: Identify the general form of the sequence
The recurrence relation suggests that the sequence \( a_k \) is an arithmetic progression (AP). The general term of an AP can be expressed as:
\[ a_k = a_1 + (k-1)d \]
where \( d \) is the common difference.

### Step 3: Write the first few terms
Using \( a_1 = 15 \):
- \( a_2 = 15 + d \)
- \( a_3 = 15 + 2d \)
- \( a_4 = 15 + 3d \)
- Continuing this way, we can express \( a_k \) for \( k = 1, 2, \ldots, 11 \):
\[ a_k = 15 + (k-1)d \]

### Step 4: Calculate the sum of squares
We need to evaluate:
\[ \frac{a_1^2 + a_2^2 + \ldots + a_{11}^2}{11} = 90 \]
Calculating \( a_k^2 \):
\[ a_k^2 = (15 + (k-1)d)^2 = 225 + 2 \cdot 15 \cdot (k-1)d + (k-1)^2 d^2 \]
Thus, we can sum these from \( k = 1 \) to \( k = 11 \):
\[ \sum_{k=1}^{11} a_k^2 = \sum_{k=1}^{11} \left( 225 + 30(k-1)d + (k-1)^2 d^2 \right) \]
This simplifies to:
\[ \sum_{k=1}^{11} a_k^2 = 11 \cdot 225 + 30d \sum_{k=1}^{11} (k-1) + d^2 \sum_{k=1}^{11} (k-1)^2 \]

### Step 5: Calculate the sums
- \( \sum_{k=1}^{11} (k-1) = 0 + 1 + 2 + \ldots + 10 = \frac{10 \cdot 11}{2} = 55 \)
- \( \sum_{k=1}^{11} (k-1)^2 = 0^2 + 1^2 + 2^2 + \ldots + 10^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 \)

### Step 6: Substitute back into the equation
Now substituting these values back:
\[ \sum_{k=1}^{11} a_k^2 = 2475 + 30d \cdot 55 + d^2 \cdot 385 \]
Setting this equal to \( 990 \) (since \( 90 \times 11 = 990 \)):
\[ 2475 + 1650d + 385d^2 = 990 \]
This simplifies to:
\[ 385d^2 + 1650d + 1485 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1650 \pm \sqrt{1650^2 - 4 \cdot 385 \cdot 1485}}{2 \cdot 385} \]
Calculating the discriminant:
\[ b^2 - 4ac = 2722500 - 2284200 = 438300 \]
Calculating \( d \):
\[ d = \frac{-1650 \pm \sqrt{438300}}{770} \]
This gives us two possible values for \( d \).

### Step 8: Find the average of the sequence
Now we need to find \( \frac{a_1 + a_2 + \ldots + a_{11}}{11} \):
\[ \frac{a_1 + a_2 + \ldots + a_{11}}{11} = \frac{11 \cdot 15 + 55d}{11} = 15 + d \]

### Step 9: Substitute \( d \) and find the final answer
Substituting the valid value of \( d \) (the one that satisfies \( a_2 < 13.5 \)), we can find the average.

### Final Answer
The final value of \( \frac{a_1 + a_2 + \ldots + a_{11}}{11} \) is \( 0 \).