JEE Problem

Using characteristic equation as mention by Challenge Master.

From the linear recurrence relation $a_k = 2a_{k-1} - a_{k-2}$ , we have the characteristic equation $x^2-2x +1 = 0$ with repeated root of $x=1$ and the general solution of $a_n = c_1 + c_2n$ .

$\implies \begin{cases} a_1 = c_1 + c_2 = 15 & ...(1) \\ a_2 = c_1 + 2c_2 & ...(2) \end{cases}$

$\begin{aligned} (2)-(1): \quad a_2 - 15 & = c_2 \\ \implies c_2 & = a_2 -15 \\ (1): \quad c_1 +a_2 -15 & = 15 \\ \implies c_1 & = 30 - a_2 \\ \implies a_n & = 30 - a_2 + n(a_2 -15) \\ & = (n-1)a_2 - 15(n-2) \end{aligned}$

From $a_n = (n-1)a_2 - 15(n-2)$ , we have:

$\begin{aligned} \sum_{n=1}^{11} a_n^2 & = \sum_{n=1}^{11} [(n-1)a_2-15(n-2)]^2 \\ & = \sum_{n=0}^{10} [na_2-15(n-1)]^2 \\ & = \sum_{n=0}^{10} [(a_2-15)^2n^2 + 30(a_2-15)n + 225] \\ & = \frac {10(11)(21)}6 \cdot (a_2-15)^2 + \frac {10(11)}2 \cdot 30(a_2-15) + 11(225) \\ & = 11 \left(35(a_2-15)^2 +150 (a_2-15) + 225 \right) \end{aligned}$

Now, we have:

$\begin{aligned} \frac 1{11} \sum_{n=1}^{11} a_n^2 & = 90 \\ 35(a_2-15)^2 +150 (a_2-15) + 225 & = 90 \\ 35(a_2-15)^2 +150 (a_2-15) + 135 & = 0 \\ 7(\color{#3D99F6}{a_2-15})^2 + 30(\color{#3D99F6}{a_2-15}) + 27 & = 0 \\ (7(\color{#3D99F6}{a_2-15})+9)(\color{#3D99F6}{a_2-15}+3) & = 0 \\ (7a_2-96)(a_2-12) & = 0 \\ \implies a_2 & = 12 \end{aligned}$

$a_2 = \dfrac {96}7$ does not satisfy the condition $27 - 2a_2 > 0$ .

Therefore, $a_n = 12(n-1) - 15(n-2) = 18-3n$ , $\implies \displaystyle \sum_{n=1}^{11} a_n = 15+12+9+6+3+0-3-6-9-12-15 = 0$ $\implies \displaystyle \frac 1{11} \sum_{n=1}^{11} a_n = \boxed{0}$ .

• From $a_k=2a_{k-1}-a_{k-2}$ , it follows that $a_k-a_{k-1}=a_{k-1}-a_{k-2}$ , which may be used to conclude that the given sequence is an Arithmetic Progression.

• Let $a-5d, a-4d, ..., a, a+d, a+2d, ..., a+5d$ be $a_1, a_2, ..., a_6, a_7, a_8, ..., a_{11}$ respectively, so that $\frac{\sum_{i=1}^{11}a_i^2}{11}$ $=90$ reduces to $a^2 + 10d^2 = 90$ . And now, we need to find $\frac{\sum_{i=1}^{11}a_i}{11}$ $=a$

• Also $a-5d=15$ , or $a=15+5d$ . Solve the equation obtained in previous step for $d$ .

• And $-2d=2a_1-2a_2=30-2a_2=3+(27-2a_2)>3$ can be used to eliminate one value of $d$ to yield the single value $d=-3$ .

• Thus, $a=15+5d=0$ .