Answer:
- x = 17; an = 3n² -3n -1
- x = 20; an = 3n² -4n +5
Step-by-step explanation:
Given the following quadratic sequences, you want the value of x and the expression for the n-th term.
Differences
One way to determine x is to look at the differences between terms. The "second difference" is constant for a quadratic sequence, and the third difference is zero.
N-th term
The quadratic equation for the n-th term can be found by solving for its coefficients. The three known values of the sequence can give rise to three linear equations in the three unknown coefficients. These can be solved by your favorite method. We use this approach in the following.
1. -1, 5, x, 35
First differences are the differences between each term and the one before:
{6, x-5, 35-x}
Second differences are the differences of these:
{x -11, 40 -2x}
Third differences are zero:
51 -3x = 0 ⇒ x = 17
The value of x is 17.
The expression for the n-th term of the sequence can be written as ...
an = a·n² +b·n +c
We are given values of a1, a2, and a4. This lets us write 3 equations for a, b, and c. The solution of those is shown on the first line of the first attachment. (The second line shows the evaluation of this quadratic equation for n=3. It gives 17, which we already knew.)
an = 3n² -3n -1
2. 4, 9, x, 37
The last line of the first attachment shows us the expression for the third differences. The value of that is zero, so ...
-3x +60 = 0 ⇒ x = 20
The value of x is 20.
As in the above problem, the matrix of equations for the quadratic coefficients can be reduced to give the coefficient values. That tells us the n-th term of this sequence is ...
an = 3n² -4n +5
The last line in the second attachment tells us this expression for the n-th term properly computes the 3rd term (x), as above.
__
Additional comments
You can also use quadratic regression to find the coefficients of the formula for the quadratic sequence. This is shown in the 3rd attachment.
If you're trying to avoid using a calculator, you can write the equations out and solve them in an ad hoc way. In case you cannot tell, the equations for the coefficients of an = a·n² +b·n +c for the first problem are ...
- 1·a +1·b +1·c = -1
- 4·a +2·b +1·c = 5
- 16·a +4·b +c = 35
You can also use the first values of the sequence (p), first difference (q), second difference (r) to write the quadratic:
an = p +(n -1)(q +(n -2)/2(r))
For (p, q, r) = (-1, 6, 6), this is an = -1 +(n -1)(3n) . . . . . . for the first sequence.
<95141404393>
