Let a1, a2, a3.....a11 be real numbers satisfying a1 = 15, 27 - 2a2 > 0 and ak = 2ak_1 – ak–2 for k = 3, 4..... 11. If a12 + a22 +.........+ a11211 = 90 , then the value of a1 + a2 + ......+ a1111 is equal to

Let a1, a2, a3.....a11 be real numbers satisfying a1 = 15, 27 - 2a2 > 0 and ak = 2ak_1 – ak–2 for k = 3, 4..... 11. If \frac{a_{1}^{2} \textrm{ }\textrm{ } + \textrm{ }\textrm{ } a_{2}^{2} \textrm{ }\textrm{ } + . ........ + \textrm{ } a_{11}^{2}}{11} \textrm{ }\textrm{ } = \textrm{ }\textrm{ } 90 , then the value of \frac{a_{1} \textrm{ }\textrm{ } + \textrm{ }\textrm{ } a_{2} \textrm{ }\textrm{ } + \textrm{ } . ..... + \textrm{ }\textrm{ } a

Engineering

Mathematics

Arithmetic Progression and Its Properties

Question

Let a₁, a₂, a₃.....a₁₁ be real numbers satisfying

a₁ = 15, 27 - 2a₂ > 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_k – a_{k – 1} = a_{k – 1} – a_{k – 1} $\Rightarrow$ a_n's are in A.P.

Let common difference d

$\frac{a_{1}^{2} + a_{2}^{2} + . .... + a_{11}^{2}}{11} = \frac{15^{2} + {(15 + d)}^{2} + . .... + {(15 + 10 d)}^{2}}{11} = 990$

on solving $\Rightarrow$ d = – 3, – 9 (rejected) $a_{2} > \frac{27}{2}$

$\frac{a_{1} + a_{2} + . .... + a_{11}}{11} = \frac{\frac{11}{2} (30 + 10 (- 3))}{11} = 0$

Understanding the Problem

We are given a sequence of real numbers a₁, a₂, ..., a₁₁ with:

a₁ = 15
27 - 2a₂ > 0
A recurrence relation: a_k = 2a_k-1 - a_k-2 for k = 3, 4, ..., 11

We are also given that the average of the squares is 90: $\frac{a_{1}^{2} + a_{2}^{2} + \dots + a_{11}^{2}}{11} = 90$

We need to find the average of the numbers themselves: $\frac{a_{1} + a_{2} + \dots + a_{11}}{11}$

Step 1: Analyze the Recurrence Relation

The recurrence relation is a_k = 2a_k-1 - a_k-2.

Let's find the characteristic equation. Assume a solution of the form a_k = r^k.

Substituting into the recurrence:

r^{k} = 2 r^{k - 1} - r^{k - 2}

Divide both sides by r^k-2 (assuming r ≠ 0):

r^{2} = 2 r - 1

Rearranging:

r^{2} - 2 r + 1 = 0

This factors as:

{(r - 1)}^{2} = 0

So, r = 1 is a repeated root. The general solution for a linear recurrence with a repeated root is an arithmetic progression (AP). The closed form is:

a_{k} = A + B (k - 1)

where A and B are constants determined by the initial conditions.

Step 2: Find the Constants A and B

We know a₁ = 15. Let's plug k=1 into our general form:

a_{1} = A + B (1 - 1) = A + B \cdot 0 = A

Therefore, A = 15.

Now, let's find a₂. We don't know its value, but we know it's related to the constant B:

a_{2} = A + B (2 - 1) = 15 + B

We are also given the inequality 27 - 2a₂ > 0. Let's use this later.

So, our sequence is defined by:

a_{k} = 15 + B (k - 1)

for k = 1, 2, ..., 11.

Step 3: Use the Given Information About the Sum of Squares

We are told:

\frac{a_{1}^{2} + a_{2}^{2} + \dots + a_{11}^{2}}{11} = 90

This means the sum of the squares is 11 * 90 = 990.

\sum_{k = 1}^{11} a_{k}^{2} = 990

We can substitute our expression for a_k into this sum:

\sum_{k = 1}^{11} {[15 + B (k - 1)]}^{2} = 990

Let's simplify the expression inside the sum. Let j = k - 1. When k=1, j=0; when k=11, j=10.

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

Expand the square:

\sum_{j = 0}^{10} (225 + 30 B j + B^{2} j^{2}) = 990

We can split this sum into three separate sums:

\sum_{j = 0}^{10} 225 + 30 B \sum_{j = 0}^{10} j + <msup

Detailed Solution

Understanding the Problem

Step 1: Analyze the Recurrence Relation

Step 2: Find the Constants A and B

Step 3: Use the Given Information About the Sum of Squares

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