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

Let $a_{1}, a_{2}, a_{3}, \cdots \cdots, a_{11}$ be real numbers satisfying $a_{1}=15,27-2 a_{2}>0$ and $a_{k}=2 a_{k-1}-a_{k-2}$ for $k=3,4, \cdots \cdots, 11 .$ If $\frac{a_{1}^{2}+a_{2}^{2}+\cdots \cdots+a_{11}^{2}}{11}=90$, then find the value of $\frac{a_{1}+a_{2}+\cdots \cdots+a_{11}}{11}$.

Key Concepts

Linear Recurrence Relations

This is a method of defining each term of a sequence based on preceding terms with a fixed rule. In the case of a second?order linear recurrence with constant coefficients, the next term is determined by a combination of the two previous terms, which often simplifies into a well-known type of sequence.

Arithmetic Sequences

An arithmetic sequence is one in which the difference between consecutive terms is constant. Recognizing that a recurrence relation produces an arithmetic sequence enables the use of formulas for the nth term and properties such as the mean of the sequence equaling the middle term when there are an odd number of terms.

Sum of Squares in an Arithmetic Sequence

When the sequence in question is arithmetic, the sum of the squares of its terms can be expressed in closed form using formulas that involve the first term and the constant difference. This sum is useful in problems where constraints are given by the average of the squares of the sequence.

Quadratic Equations

These arise when substituting the arithmetic sequence expressions into given conditions, leading to an equation in the common difference or other parameters. Solving the quadratic equation determines the values required to satisfy all constraints in the problem.

Transcript

00:01 In this question we have to find the value of a1 plus a2 up to a11 upon 11.

00:08 Given ak is equal to 2 into a, k minus 1 minus a k minus 2, 4, k is equal to 3, 4, 2 up to 11.

00:48 2 into a, k minus 1, k minus 2.

00:51 Is equal to ak plus a k minus 2 therefore a k minus 2 comma a k minus 2 comma a k minus 1 comma ak are in ap 4 k is equal to 3 4 up to 11 therefore a 1 a2, a3 up to a11 are in ap.

02:06 Let d is equal to common difference of ap.

02:30 Given a1 is equal to 15 and a1 square plus a2 square plus a3 square up to a 11 square upon 11 is equal to 90 summation k is equal to 1 to 11 a k square is equal to 11 minus 1 into d whole to the power 2 is equal to 990.

04:07 Summation k is equal to 1 to 11, 21, 225 plus 30 into k minus 1 into d plus k minus 1 whole square into d square is equal to 990 summation k is equal to 1 to 11 225 plus 30 d into summation k is equal to 1 to 11 to 11 k minus 1 plus d square into summation summation k is equal to 1 to 11 k minus 1 whole square is equal to 990 is equal to 225 into 1 plus 1 plus 11 times into 30 d into 1 plus 2 plus 3 up to 10 plus d square into 1 square plus 2 square plus 3 square up to 10 square is equal to 990 is equal to 225 into 11 plus 30 d into we know that sum of first n natural number is equal n into n plus 1 upon 2 here n is equal to 10 and sum of a square of first n natural number is n into n plus 1 into 2 n plus 1 upon 6 here n is 10 30 d into 10 10 into 10 into 11 upon 2 plus d square into 10 into 11 into 21 upon 6 is equal to 990 taking common 11 225 plus 150 plus 35d square is equal to 990.

09:21 Divide both sides by 11.

09:27 35 d square plus 150d plus 225 is equal to 90.

09:48 35 d square plus 150 d plus 150 d plus 1 ,000 35 is equal to 90.

10:09 Taking common 5, 7d square plus 30 d plus 27 is equal to 0...

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