Question 1108399: -3, -2, x, -6, y................is a quadratic sequence.
determine the value of x and y.
Found 2 solutions by rothauserc, greenestamps:
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! general form of quadratic sequence is
:
an^2 + bn + c, where a,b,c are real numbers
:
for n=1, a + b + c = -3
for n=2, 4a + 2b + c = -2
for n=3, 9a + 3b + c = x
for n=4, 16a + 4b + c = -6
for n=5, 25a + 5b + c = y
:
we will use the following 3 equations with 3 unknowns
:
a + b + c = -3
4a + 2b + c = -2
16a + 4b + c = -6
:
use any linear equation solver for 3 equations with 3 unknows
:
a = -1, b = 4, c = -6
:
the equation for the nth term is
:
-n^2 +4b -6
:
for n = 1, we have -1 +4 -6 = -3
for n = 2, we have -4 +8 -6 = -2
for n = 3, we have -9 +12 -6 = -3
for n = 4, we have -16 +16 -6 = -6
for n = 5, we have -25 +20 -6 = -11
:
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x = -3 and y = -11
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:
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The solution by the other tutor is a perfectly good one, using a standard algebraic technique. Here is another method that is often useful on problems like this.
In any quadratic sequence, the row of "second differences" is constant. Here is an example.
For f(x) = 2x^2+3x-4, the first few terms of the sequence are
1, 10, 23, 40, 61
The "first differences" (differences between successive terms of the sequence) ar
9, 13, 17, 21
The "second differences" (differences between successive first differences) are
4, 4, 4
This will always be the case for a quadratic sequence.
So we can apply this knowledge to solving your problem.
The terms of your sequence are
-3, -2, x, -6, y
The first differences are
1, x+2, -6-x, y+6
The second differences are
x+1, -2x-8, x+y+12
Since we know the second differences are constant, we have
Then
As found by the other tutor, the missing numbers are x = -3 and y = -11.
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