Algebra Examples

2 + 4 + 6 + 8

$2 + 4 + 6 + 8$

Step 1

This is the formula to find the sum of the first

n

$n$ terms of the sequence. To evaluate it, the values of the first and

n

$n$ th terms must be found.

S_{n} = \frac{n}{2} \cdot (a_{1} + a_{n})

$S_{n} = \frac{n}{2} \cdot (a_{1} + a_{n})$

Step 2

This is an arithmetic sequence since there is a common difference between each term. In this case, adding

2

$2$ to the previous term in the sequence gives the next term. In other words,

a_{n} = a_{1} + d (n - 1)

$a_{n} = a_{1} + d (n - 1)$ .

Arithmetic Sequence:

d = 2

$d = 2$

Step 3

This is the formula of an arithmetic sequence.

a_{n} = a_{1} + d (n - 1)

$a_{n} = a_{1} + d (n - 1)$

Step 4

Substitute in the values of

a_{1} = 2

$a_{1} = 2$ and

d = 2

$d = 2$ .

a_{n} = 2 + 2 (n - 1)

$a_{n} = 2 + 2 (n - 1)$

Step 5

Simplify each term.

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Apply the distributive property.

a_{n} = 2 + 2 n + 2 \cdot - 1

$a_{n} = 2 + 2 n + 2 \cdot - 1$

Multiply

2

$2$ by

- 1

$- 1$ .

a_{n} = 2 + 2 n - 2

$a_{n} = 2 + 2 n - 2$

a_{n} = 2 + 2 n - 2

$a_{n} = 2 + 2 n - 2$

Step 6

Combine the opposite terms in

2 + 2 n - 2

$2 + 2 n - 2$ .

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Subtract

2

$2$ from

2

$2$ .

a_{n} = 2 n + 0

$a_{n} = 2 n + 0$

Add

2 n

$2 n$ and

0

$0$ .

a_{n} = 2 n

$a_{n} = 2 n$

a_{n} = 2 n

$a_{n} = 2 n$

Step 7

Substitute in the value of

n

$n$ to find the

n

$n$ th term.

a_{4} = 2 (4)

$a_{4} = 2 (4)$

Step 8

Multiply

2

$2$ by

4

$4$ .

a_{4} = 8

$a_{4} = 8$

Step 9

Replace the variables with the known values to find

S_{4}

$S_{4}$ .

S_{4} = \frac{4}{2} \cdot (2 + 8)

$S_{4} = \frac{4}{2} \cdot (2 + 8)$

Step 10

Divide

4

$4$ by

2

$2$ .

S_{4} = 2 \cdot (2 + 8)

$S_{4} = 2 \cdot (2 + 8)$

Step 11

Add

2

$2$ and

8

$8$ .

S_{4} = 2 \cdot 10

$S_{4} = 2 \cdot 10$

Step 12

Multiply

2

$2$ by

10

$10$ .

S_{4} = 20

$S_{4} = 20$

Step 13

Convert the fraction to a decimal.

S_{4} = 20

$S_{4} = 20$