Let $a_1, a_2, a_3, \ldots, a_{11}$ be real numbers satisfying $a_1 = 15$, $27 - 2a_2 > 0$ and $a_k = 2a_{k-1} - a_{k-2}$ for $k = 3, 4, \ldots, 11$. if $\frac{a_1^2 + a_2^2 + \cdots + a_{11}^2}{11} = 90$, then the value of $\frac{a_1 + a_2 + \cdots + a_{11}}{11}$ is equal to

Let a₁, a₂, a₃, …, a₁₁ be real numbers satisfying a₁ = 15, 27 - 2a₂ > 0 and aₖ = 2aₖ₋₁ - aₖ₋₂ for k = 3, 4, …, 11. if (a₁² + a₂² + … + a₁₁²)/11 = 90, then the value of (a₁ + a₂ + … + a₁₁)/11 is equal to $a_1 = 15$ $a_k = 2a_{k-1} - a_{k-2}, \quad k = 3, 4, \ldots, 11$ $27 - 2a_2 > 0$ $\frac{a_1^2 + a_2^2 + \ldots + a_{11}^2}{11} = 90$ $\text{Find: } \frac{a_1 + a_2 + \ldots + a_{11}}{11}$ $a_k - a_{k-1} = a_{k-1} - a_{k-2}$ $\therefore a_k = a_1 + (k-1)d$ $a_2 = a_1 + d = 15 + d$ $27 - 2a_2 > 0$ $\Rightarrow 27 - 2(15+d) > 0$ $\Rightarrow -3 - 2d > 0$ $\Rightarrow 2d < -3$ $d < -\frac{3}{2}$ $\sum_{k=1}^{11} a_k^2 = 11 \times 90$ $= 990$ $\sum_{k=1}^{11} \left(15 + (k-1)d\right)^2 = 990$ $= 225\sum 1 + 30d\sum (k-1) + d^2\sum (k-1)^2$ $\sum 1 = 11, \quad \sum_{k=1}^{11}(k-1) = 55, \quad \sum (k-1)^2 = 385$ $= 2475 + 1650d + 385d^2$ $\Rightarrow 385d^2 + 1650d + 1485 = 0$ $\Rightarrow 7d^2 + 30d + 27 = 0$ $D = 30^2 - 4 \cdot 7 \cdot 27$ $= 900 - 756 = 144$ $\sqrt{D} = 12$ $d = \frac{-30 \pm 12}{14}$ $d = -\frac{9}{7} \quad \text{or} \quad d = -3$ $d < -\frac{3}{2}$ $-\frac{9}{7} > -\frac{3}{2} \ \Rightarrow \ \text{reject}$ $-3 < -\frac{3}{2} \ \Rightarrow \ \text{accept}$ $\frac{1}{11}\sum_{k=1}^{11} a_k = 15 + \frac{55d}{11}$ $= 15 + 5d$ $= 15 + 5(-3)$ $= 0$ $15, 12, 9, 6, 3, 0, -3, -6, -9, -12, -15$ $\sum a_k = 0, \quad \sum a_k^2 = 2(15^2+12^2+9^2+6^2+3^2) = 990$ $\therefore \frac{a_1 + a_2 + \ldots + a_{11}}{11} = 0$