Let $a_{1}, a_{2}, a_{3}, \ldots . . ., a_{11}$ be real numbers satisfying $a_{1}=15,27-2 a_{2}0$ a

Question

Let $a_{1}, a_{2}, a_{3}, \ldots . . ., a_{11}$ be real numbers satisfying $a_{1}=15,27-2 a_{2}>0$ and $a_{k}=2 a_{k-1}-a_{k-2}$ 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

Professor PC Thomas Classes · Answer

(0) : $a_{k}=2 a_{k-1}-a_{k-2}$\ $\Rightarrow \quad a_{k}-a_{k-1}=a_{k-1}-a_{k-2}$ $=a_{k-2}-a_{k-3}=\ldots$\ Thus the sequence $\left\{a_{k}ight\}$ is an A.P. $$ \begin{aligned} & \frac{a_{1}^{2}+a_{2}^{2}+\ldots+a_{11}^{2}}{11}=90 \ \Rightarrow & \frac{15^{2}+(15+	heta)^{2}+\ldots+(15+10 	heta)^{2}}{11}=90 \ \Rightarrow & 7 	heta^{2}+30 	heta+27=0 \Rightarrow(7 	heta+9)(	heta+3)=0 \end{aligned} $$ But $27-2 a_{2}>0 \Rightarrow a_{2}<27 / 2 	herefore 	heta=-3$\ $	herefore \quad \frac{a_{1}+a_{2}+\ldots+a_{11}}{11}=\frac{11}{2} \cdot \frac{\{30+10(-3)\}}{11}=0$\