Find Mean of Eleven Terms in Arithmetic Progression

To find the mean of the eleven terms, $\frac{a_1 + a_2 + \dots + a_{11}}{11}$, we first analyze the given conditions.

The recurrence relation is given as:
$a_k = 2a_{k-1} - a_{k-2} \implies a_k - a_{k-1} = a_{k-1} - a_{k-2} \quad \text{for } k = 3, 4, \dots, 11$

This implies that the sequence $a_1, a_2, \dots, a_{11}$ forms an Arithmetic Progression (A.P.). Let $d$ be the common difference of this A.P., and we are given $a_1 = 15$.
Therefore, the $k$-th term of the A.P. is:
$a_k = 15 + (k-1)d$

We are given the inequality $27 - 2a_2 > 0$. Substituting $a_2 = 15 + d$:
$27 - 2(15 + d) > 0 \implies 27 - 30 - 2d > 0 \implies -3 - 2d > 0 \implies d < -\frac{3}{2}$

Let $\bar{a}$ denote the mean of the 11 terms. Since the terms are in A.P., the mean is equal to the middle term ($a_6$):
$\bar{a} = \frac{a_1 + a_2 + \dots + a_{11}}{11} = a_6 = 15 + 5d$

From this, we can express $d$ in terms of $\bar{a}$:
$d = \frac{\bar{a} - 15}{5}$

Next, we know that the variance of $n$ terms of an A.P. with common difference $d$ is given by the formula:
$\text{Variance} = \frac{n^2 - 1}{12} d^2$

For $n = 11$:
$\text{Variance} = \frac{11^2 - 1}{12} d^2 = \frac{120}{12} d^2 = 10d^2$

Alternatively, by definition, the variance is also equal to:
$\text{Variance} = \frac{a_1^2 + a_2^2 + \dots + a_{11}^2}{11} - \bar{a}^2$

Given that $\frac{a_1^2 + a_2^2 + \dots + a_{11}^2}{11} = 90$, we get:
$\text{Variance} = 90 - \bar{a}^2$

Equating the two expressions for variance:
$90 - \bar{a}^2 = 10d^2 \implies \bar{a}^2 + 10d^2 = 90$

Substitute $d = \frac{\bar{a} - 15}{5}$ into the equation:
$\bar{a}^2 + 10 \left( \frac{\bar{a} - 15}{5} \right)^2 = 90$
$\bar{a}^2 + \frac{2}{5} (\bar{a} - 15)^2 = 90$

Multiplying the entire equation by 5:
$5\bar{a}^2 + 2(\bar{a}^2 - 30\bar{a} + 225) = 450$
$7\bar{a}^2 - 60\bar{a} + 450 = 450$
$7\bar{a}^2 - 60\bar{a} = 0$
$\bar{a}(7\bar{a} - 60) = 0$

This gives two possible solutions for $\bar{a}$:

1. If $\bar{a} = 0$:
   $d = \frac{0 - 15}{5} = -3$
   Checking the condition $d < -\frac{3}{2}$: $-3 < -1.5$, which is satisfied.

2. If $\bar{a} = \frac{60}{7}$:
   $d = \frac{\frac{60}{7} - 15}{5} = -\frac{9}{7} \approx -1.286$
   Checking the condition $d < -\frac{3}{2}$: $-\frac{9}{7} > -1.5$, which violates the given inequality.

Thus, the only valid value for the mean is $\bar{a} = 0$.

Therefore, the value of $\frac{a_1 + a_2 + \dots + a_{11}}{11}$ is equal to $0$.