Let a_1,a_2,a_3,,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,,11. If a_1^2 + a_2^2 + + a_11^211 = 90, then the value of a_1 + a_2 + + a_1111 is equal

aligned& a_k=2 a_k-1-a_k-2 \\& a_k-a_k-1=a_k-1-a_k-2=a_k-2-a_k-3=aligned Thus the sequence \a_k\ is an A.P.aligned& a_1^2+a_2^2++a_11^211=90 \\& 15^2+(15+)^2++(15+10 )^211=90 \\& 7 ^2+30 +27=0 (7 +9)(+3)=

Let a_{1},a_{2},a_{3},,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,,11. If \frac{a_{1}^{2} + a_{2}^{2} + + a_{11}^{2}}{11} = 90, then the value of \frac{a_{1} + a_{2} + + a_{11}}{11} is equal: Sequence & Series (Mathematics)

$\begin{aligned}& a_k=2 a_{k-1}-a_{k-2} \\& \Rightarrow a_k-a_{k-1}=a_{k-1}-a_{k-2}=a_{k-2}-a_{k-3}=\cdots\end{aligned}$ Thus the sequence $\left\{a_k\right\}$ is an A.P. $\begin{aligned}& \frac{a_1^2+a_2^2+\cdots+a_{11}^2}{11}=90 \\& \Rightarrow \frac{15^2+(15+\theta)^2+\cdots+(15+10 \theta)^2}{11}=90 \\& \Rightarrow 7 \theta^2+30 \theta+27=0 \Rightarrow(7 \theta+9)(\theta+3)=0\end{aligned}$ But $27-2 a_2>0 \Rightarrow a_2<27 / 2 \therefore \theta=-3$ Now $\frac{a_1+a_2+\cdots+a_{11}}{11}=\frac{11}{2} \cdot \frac{\{30+10(-3)\}}{11}=0$

Explanation

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