Question

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 $\frac{a_1^2+a_2^2+a_3^2+....a_{11}^2}{11}=90$, then $\frac{a_1+a_2+a_3+.....a_{11}}{11}$ is equal to

• A. 4
• B. 6
• C. 9
• D. 0

Answer 1

The correct option is D

0

$a_{k-1}=\frac{a_k+a_{k-2}}{2}\Rightarrow a_1,a_2,....a_{11}$ are in A.P.

Let d be the common difference of A.P.

${\sum }_{r=0}^{10}(15+rd{)}^2=990\Rightarrow {15}^2\times 11+d^2\times \frac{10\times 11\times 21}{6}+30d\times \frac{10\times 11}{2}=990$

$\Rightarrow 7d^2+30d+27=0\Rightarrow d=-3$ or $-\frac{9}{7}$

$a_2=15+d=12$ or $\frac{96}{7}$. But $a_2<\frac{27}{2}\Rightarrow d=-3$

$S_{11}=\frac{11}{2}(2\times 15+10\times -3)=0$