Precalculus Examples

$10$ ,

$8$ ,

$5$ ,

$1$ ,

$- 4$

Step 1

Find the first level differences by finding the differences between consecutive terms.

$- 2, - 3, - 4, - 5$

Step 2

Find the second level difference by finding the differences between the first level differences. Because the second level difference is constant, the sequence is quadratic and given by

$a_{n} = a n^{2} + b n + c$ .

$- 1$

Step 3

Solve for

$a$ by setting

$2 a$ equal to the constant second level difference

$- 1$ .

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Step 3.1

Set

$2 a$ equal to the constant second level difference

$- 1$ .

$2 a = - 1$

Step 3.2

Divide each term in

$2 a = - 1$ by

$2$ and simplify.

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Step 3.2.1

Divide each term in

$2 a = - 1$ by

$2$ .

$\frac{2 a}{2} = \frac{- 1}{2}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of

$2$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{2 a}{2} = \frac{- 1}{2}$

Step 3.2.2.1.2

Divide

$a$ by

$1$ .

$a = \frac{- 1}{2}$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Move the negative in front of the fraction.

$a = - \frac{1}{2}$

Step 4

Solve for

$b$ by setting

$3 a + b$ equal to the first first level difference

$- 2$ .

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Step 4.1

Set

$3 a + b$ equal to the first first level difference

$- 2$ .

$3 a + b = - 2$

Step 4.2

Substitute

$- \frac{1}{2}$ for

$a$ .

$3 (- \frac{1}{2}) + b = - 2$

Step 4.3

Simplify each term.

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Step 4.3.1

Multiply

$3 (- \frac{1}{2})$ .

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Step 4.3.1.1

Multiply

$- 1$ by

$3$ .

$- 3 (\frac{1}{2}) + b = - 2$

Step 4.3.1.2

Combine

$- 3$ and

$\frac{1}{2}$ .

$\frac{- 3}{2} + b = - 2$

Step 4.3.2

Move the negative in front of the fraction.

$- \frac{3}{2} + b = - 2$

Step 4.4

Move all terms not containing

$b$ to the right side of the equation.

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Step 4.4.1

Add

$\frac{3}{2}$ to both sides of the equation.

$b = - 2 + \frac{3}{2}$

Step 4.4.2

To write

$- 2$ as a fraction with a common denominator, multiply by

$\frac{2}{2}$ .

$b = - 2 \cdot \frac{2}{2} + \frac{3}{2}$

Step 4.4.3

Combine

$- 2$ and

$\frac{2}{2}$ .

$b = \frac{- 2 \cdot 2}{2} + \frac{3}{2}$

Step 4.4.4

Combine the numerators over the common denominator.

$b = \frac{- 2 \cdot 2 + 3}{2}$

Step 4.4.5

Simplify the numerator.

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Step 4.4.5.1

Multiply

$- 2$ by

$2$ .

$b = \frac{- 4 + 3}{2}$

Step 4.4.5.2

Add

$- 4$ and

$3$ .

$b = \frac{- 1}{2}$

Step 4.4.6

Move the negative in front of the fraction.

$b = - \frac{1}{2}$

Step 5

Solve for

$c$ by setting

$a + b + c$ equal to the first term in the sequence

$10$ .

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Step 5.1

Set

$a + b + c$ equal to the first term in the sequence

$10$ .

$a + b + c = 10$

Step 5.2

Substitute

$- \frac{1}{2}$ for

$a$ and

$- \frac{1}{2}$ for

$b$ .

$- \frac{1}{2} - \frac{1}{2} + c = 10$

Step 5.3

Simplify

$- \frac{1}{2} - \frac{1}{2} + c$ .

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Step 5.3.1

Combine the numerators over the common denominator.

$c + \frac{- 1 - 1}{2} = 10$

Step 5.3.2

Simplify the expression.

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Step 5.3.2.1

Subtract

$1$ from

$- 1$ .

$c + \frac{- 2}{2} = 10$

Step 5.3.2.2

Divide

$- 2$ by

$2$ .

$c - 1 = 10$

Step 5.4

Move all terms not containing

$c$ to the right side of the equation.

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Step 5.4.1

Add

$1$ to both sides of the equation.

$c = 10 + 1$

Step 5.4.2

Add

$10$ and

$1$ .

$c = 11$

Step 6

Substitute the values of

$a$ ,

$b$ , and

$c$ into the quadratic sequence formula

$a_{n} = a n^{2} + b n + c$ .

$a_{n} = - \frac{1}{2} n^{2} - \frac{1}{2} n + 11$

Step 7

Simplify each term.

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Step 7.1

Combine

$n^{2}$ and

$\frac{1}{2}$ .

$a_{n} = - \frac{n^{2}}{2} - \frac{1}{2} n + 11$

Step 7.2

Combine

$n$ and

$\frac{1}{2}$ .

$a_{n} = - \frac{n^{2}}{2} - \frac{n}{2} + 11$