Question

Let $a_1,a_2,.........a_{11}$ 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 to

• A. $0$
• B. $1$
• C. $11$
• D. $100$

Answer 1

The correct option is A $0$

$a_1=15$
$\frac{a_k+a_{k-2}}{2}=a_{k-1}fork=3,4,.........11$
$\Rightarrow a_1,a_2,a_3,............a_{11}$ are in A.P.
$\frac{a_1^2+a_2^2+........+a_{11}^2}{11}=90$
$\Rightarrow 7d^2+30d+27=0\Rightarrow d=-3or-\frac{9}{7}$
Since $27-2a_2>0\Rightarrow a_2<\frac{27}{2}\Rightarrow d=-3$
$\frac{a_1+a_2+..........+a_{11}}{11}=\frac{11}{2}\frac{\left[\right(30+10(-3)]}{11}=0$