Find the 6th Term 1 , 3 , 6 , 10 , 15

$1$ , $3$ , $6$ , $10$ , $15$

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 $1$ .

Set $2 a$ equal to the constant second level difference $1$ .

$2 a = 1$

Divide each term in $2 a = 1$ by $2$ and simplify.

Divide each term in $2 a = 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $a$ by $1$ .

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

Step 4

Solve for $b$ by setting $3 a + b$ equal to the first first level difference $2$ .

Set $3 a + b$ equal to the first first level difference $2$ .

$3 a + b = 2$

Substitute $\frac{1}{2}$ for $a$ .

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

Combine $3$ and $\frac{1}{2}$ .

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

Move all terms not containing $b$ to the right side of the equation.

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

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

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Combine $2$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $2$ by $2$ .

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

Subtract $3$ from $4$ .

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

Step 5

Solve for $c$ by setting $a + b + c$ equal to the first term in the sequence $1$ .

Set $a + b + c$ equal to the first term in the sequence $1$ .

$a + b + c = 1$

Substitute $\frac{1}{2}$ for $a$ and $\frac{1}{2}$ for $b$ .

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

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

Combine the numerators over the common denominator.

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

Simplify the expression.

Add $1$ and $1$ .

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

Divide $2$ by $2$ .

$c + 1 = 1$

Move all terms not containing $c$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$c = 1 - 1$

Subtract $1$ from $1$ .

$c = 0$

Step 6

Substitute the values of $a$ , $b$ , and $c$ into the quadratic sequence formula $a_{n} = a n^{2} + b n + c$ .

$a_{n} = \frac{1}{2} n^{2} + \frac{1}{2} n + 0$

Step 7

Simplify.

Add $\frac{1}{2} n^{2} + \frac{1}{2} n$ and $0$ .

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

Simplify each term.

Combine $\frac{1}{2}$ and $n^{2}$ .

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

Combine $\frac{1}{2}$ and $n$ .

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

Step 8

Substitute in the value of $n$ to find the $n$ th term.

$a_{6} = \frac{{(6)}^{2}}{2} + \frac{6}{2}$

Step 9

Combine the numerators over the common denominator.

$a_{6} = \frac{{(6)}^{2} + 6}{2}$

Step 10

Simplify the expression.

Raise $6$ to the power of $2$ .

$a_{6} = \frac{36 + 6}{2}$

Add $36$ and $6$ .

$a_{6} = \frac{42}{2}$

Divide $42$ by $2$ .

$a_{6} = 21$

Find the 6th Term 1 , 3 , 6 , 10 , 15

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