Question

Let $a_1,a_2,a_3,\dots ,a_{11}$ be real numbers satisfying $a_1=15,$ $27-2a_2>0$ and $a_k=2a_{k-1}-a_{k-2}$ for $k=3,4,\dots ,11.$ If $\frac{(a_1{)}^2+(a_2{)}^2+\cdots +(a_{11}{)}^2}{11}=90,$ then the value of $a_5$ is

• A. $3$
• B. $5$
• C. $12$
• D. $14$

Answer 1

The correct option is A $3$

$a_k=2a_{k-1}-a_{k-2}$
$\Rightarrow a_k+a_{k-2}=2a_{k-1}$
$\Rightarrow a_1,a_2,a_3,\dots ,a_{11}$ are in A.P.
Let the common difference be $d.$
$a_1=15,a_2=15+d,a_3=15+2d,\dots ,a_{11}=15+10d$

Now, $\frac{(a_1{)}^2+(a_2{)}^2+\cdots +(a_{11}{)}^2}{11}$
$=\frac{(15{)}^2+(15+d{)}^2+(15+2d{)}^2+\cdots +(15+10d{)}^2}{11}=90$
$\Rightarrow \frac{(15{)}^2\cdot 11+30d(1+2+\cdots +10)+d^2(1^2+2^2+\cdots +{10}^2)}{11}=90$
$\Rightarrow 35d^2+150d+225=90$
$\Rightarrow 35d^2+150d+135=0$
$\Rightarrow 7d^2+30d+27=0$
$\Rightarrow 7d^2+21d+9d+27=0$
$\Rightarrow (d+3)(7d+9)=0$
$\Rightarrow d=-3,\frac{-9}{7}$

For $d=\frac{-9}{7},$
$a_2=15-\frac{9}{7}=\frac{96}{7}>\frac{27}{2}$
Therefore, $d\ne \frac{-9}{7}$
$\therefore d=-3$
Hence, $a_5=15+4(-3)=3$