Let a1, a2, a3, ..., a11 be real numbers satisfying a1=15, 27-2a20 and ak=2ak-1-ak-2 for k=3, 4, ..

Q: Wait, how do I know for sure this is an Arithmetic Progression? Can't it be something else?

That is a great question! The recurrence $a_k = 2a_{k-1} - a_{k-2}$ is a linear homogeneous recurrence relation with constant coefficients. When you rearrange it to $a_k - a_{k-1} = a_{k-1} - a_{k-2}$, you are explicitly stating that the difference between consecutive terms is constant. By definition, any sequence with a constant difference between consecutive terms is an Arithmetic Progression. There is no other possibility!

Q: I got two values for $d$, $d = -3$ and $d = -9/7$. Why do I have to reject one?

In JEE problems, constraints are not just suggestions; they are the boundary conditions of the universe we are working in. The problem explicitly states $27 - 2a_2 > 0$. Since $a_2 = a_1 + d = 15 + d$, this inequality becomes $27 - 2(15 + d) > 0$, which simplifies to $d < -1.5$. While $-3$ satisfies this, $-9/7$ (which is approximately $-1.28$) does not. Always check your roots against the given constraints!

Q: Is there a faster way to find the mean of the sequence without calculating all the terms?

Absolutely! For any Arithmetic Progression with an odd number of terms, the arithmetic mean is always equal to the middle term. Since we have $11$ terms, the middle term is the $6$th term, $a_6$. Once you have $a_1$ and $d$, you can simply calculate $a_6 = a_1 + 5d$. It is much faster and less prone to calculation errors than summing everything up.

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

Sigma Percentile · Accepted Answer

### The Hidden Geometry of Sequences

Have you ever looked at a sequence of numbers and felt like they were just a random collection of digits? In the world of JEE Advanced, nothing is ever truly random. Every recurrence relation is a story waiting to be told.

Today, we are going to decode the story of $a_1, a_2, \dots, a_{11}$.

### Phase 1: Decoding the Recurrence

We are given the recurrence $a_k = 2a_{k-1} - a_{k-2}$. At first glance, this might look like a standard recursive definition.

If we rearrange this equation, we get $a_k - a_{k-1} = a_{k-1} - a_{k-2}$. This is the "Aha!" moment.

This equation tells us that the gap between the $k$-th term and the $(k-1)$-th term is exactly the same as the gap between the $(k-1)$-th term and the $(k-2)$-th term. This is the very definition of an **Arithmetic Progression (A.P.)**.

We have a first term $a_1 = 15$ and a common difference $d$. Our sequence is simply $15, 15+d, 15+2d, \dots, 15+10d$.

### Phase 2: The Algebraic Grind

We are told that the mean of the squares of these eleven terms is $90$. Mathematically, this is expressed as:

$$
\frac{1}{11} \sum_{i=1}^{11} a_i^2 = 90 \Rightarrow \sum_{i=1}^{11} a_i^2 = 990
$$

To handle this, let us use the general term $a_i = 15 + (i-1)d$. To make our lives easier, let us shift the index by setting $j = i-1$.

Now, our summation becomes:

$$
\sum_{j=0}^{10} (15 + jd)^2 = 990
$$

Expanding this, we get $\sum_{j=0}^{10} (225 + 30jd + j^2d^2) = 990$. By distributing the summation, we obtain:

$$
225 \sum_{j=0}^{10} 1 + 30d \sum_{j=0}^{10} j + d^2 \sum_{j=0}^{10} j^2 = 990
$$

### Phase 3: The Quadratic Battle

Now, we use our standard summation formulas. The sum of $1$ from $j=0$ to $10$ is $11$. The sum of $j$ from $0$ to $10$ is $\frac{10 	imes 11}{2} = 55$.

The sum of $j^2$ from $0$ to $10$ is:

$$
\frac{10 	imes 11 	imes 21}{6} = 385
$$

Plugging these in, we get $225(11) + 30d(55) + 385d^2 = 990$. This simplifies to $2475 + 1650d + 385d^2 = 990$.

Rearranging, we arrive at the quadratic:

$$
385d^2 + 1650d + 1485 = 0
$$

Dividing by $55$, we get $7d^2 + 30d + 27 = 0$. Factoring this, we find $(7d + 9)(d + 3) = 0$, giving us $d = -3$ or $d = -9/7$.

### Phase 4: The Gatekeeper

We have two candidates for $d$. But we must respect the constraint $27 - 2a_2 > 0$.

Since $a_2 = 15 + d$, this becomes $27 - 2(15 + d) > 0$, or $-3 - 2d > 0$, which means $d < -1.5$.

Our candidate $d = -3$ is clearly less than $-1.5$, but $d = -9/7 \approx -1.28$ is not. Thus, we must reject $-9/7$ and accept **$d = -3$**.

### Phase 5: The Elegant Finish

Finally, we need the mean of the sequence: $\frac{a_1 + a_2 + \dots + a_{11}}{11}$. In an A.P. with an odd number of terms, the mean is simply the middle term, $a_6$.

Since $a_6 = a_1 + 5d$, we calculate:

$$
15 + 5(-3) = 15 - 15 = 0
$$

The symmetry of the sequence around the middle term leads us to a beautiful, clean result of **$0$**. You have conquered the problem!

Enter Numerical Value:

Visualized Solution

Identifying the Sequence Type

Setting up the Sum of Squares

Expanding the Summation

Expanding the Square

Using Summation Formulas

Forming the Quadratic Equation

Solving for $d$

Applying the Constraint

Selecting the Valid $d$

Final Calculation

The Sigma Insight: Arithmetic Progression (A.P.)

The Hidden Geometry of Sequences

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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