Let a_1, a_2, a_3, ldots, a_{11} be real numb | vedclass

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(A) The recurrence relation $a_k = 2a_{k-1} - a_{k-2}$ implies that $a_1, a_2, \ldots, a_{11}$ form an Arithmetic Progression ($A$.$P$.).Let the first

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(A) The recurrence relation $a_k = 2a_{k-1} - a_{k-2}$ implies that $a_1, a_2, \ldots, a_{11}$ form an Arithmetic Progression ($A$.$P$.).Let the first term be $a = a_1 = 15$ and the common difference be $d$.Then $a_k = a + (k-1)d$.The sum of squares is $\sum_{k=1}^{11} a_k^2 = \sum_{k=0}^{10} (a + kd)^2 = 11a^2 + 2ad \sum_{k=0}^{10} k + d^2 \sum_{k=0}^{10} k^2$.Using $\sum k = 55$ and $\sum k^2 = 385$,we get $\frac{11a^2 + 110ad + 385d^2}{11} = a^2 + 10ad + 35d^2 = 90$.Substituting $a = 15$: $225 + 150d + 35d^2 = 90$,which simplifies to $35d^2 + 150d + 135 = 0$.Dividing by $5$: $7d^2 + 30d + 27 = 0 \Rightarrow (7d + 9)(d + 3) = 0$.So,$d = -3$ or $d = -9/7$.Given $27 - 2a_2 > 0$,where $a_2 = a + d = 15 + d$,we have $27 - 2(15 + d) > 0$ $\Rightarrow 27 - 30 - 2d > 0$ $\Rightarrow -3 - 2d > 0$ $\Rightarrow d Thus,$d = -3$ is the only valid solution.The mean is $\frac{a_1 + \ldots + a_{11}}{11} = a + 5d = 15 + 5(-3) = 0$.

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 $\frac{a_1^2 + a_2^2 + \ldots + a_{11}^2}{11} = 90$,then the value of $\frac{a_1 + a_2 + \ldots + a_{11}}{11}$ is equal to

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