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Solution - Arithmetic sequences

The common difference equals:

130

130

The sum of the sequence equals:

507

507

The explicit formula of this sequence is:

a_{n} = 39 + (n - 1) \cdot 130

a_n=39+(n-1)*130

The recursive formula of this sequence is:

a_{n} = a_{(n - 1)} + 130

a_n=a_((n-1))+130

The nth terms:

39, 169, 299, 429, 559, 689...

39,169,299,429,559,689...

See steps

Step-by-step explanation

1. Find the common difference

Find the common difference by subtracting any term in the sequence from the term that comes after it.

$a_{2} - a_{1} = 169 - 39 = 130$

$a_{3} - a_{2} = 299 - 169 = 130$

The difference of the sequence is constant and equals the difference between two consecutive terms.
$d = 130$

2. Find the sum

Calculate the sum of the sequence using the sum formula:

$S u m = (n (a_{1} + a_{n})) / 2$

$S u m = (n * (a_{1} + a_{n})) / 2$

Plug in the terms.

$S u m = (3 * (a_{1} + a_{n})) / 2$

$S u m = (3 * (39 + a_{n})) / 2$

$S u m = (3 * (39 + 299)) / 2$

Simplify the expression.

$S u m = (3 * (39 + 299)) / 2$

$S u m = (3 * 338) / 2$

$S u m = \frac{1014}{2}$

$S u m = 507$

The sum of this sequence is $507$ .

This series corresponds to the following straight line $y = 130 x + 39$

3. Find the explicit form

The formula for expressing arithmetic sequences in their explicit form is:
$a_{n} = a_{1} + (n - 1) \cdot d$

Plug in the terms.
$a_{1} = 39$ (this is the 1st term)
$d = 130$ (this is the common difference)
$a_{n}$ (this is the nth term)
$n$ (this is the term position)

The explicit form of this arithmetic sequence is:

$a_{n} = 39 + (n - 1) \cdot 130$

4. Find the recursive form

The formula for expressing arithmetic sequences in their recursive form is:
$a_{n} = a_{(1 - n)} + d$

Plug in the d term.
$d = 130$ (this is the common difference)

The recursive form of this arithmetic sequence is:

$a_{n} = a_{(n - 1)} + 130$

5. Find the nth element

$a_{1} = a_{1} + (n - 1) \cdot d = 39 + (1 - 1) \cdot 130 = 39$

$a_{2} = a_{1} + (n - 1) \cdot d = 39 + (2 - 1) \cdot 130 = 169$

$a_{3} = a_{1} + (n - 1) \cdot d = 39 + (3 - 1) \cdot 130 = 299$

$a_{4} = a_{1} + (n - 1) \cdot d = 39 + (4 - 1) \cdot 130 = 429$

$a_{5} = a_{1} + (n - 1) \cdot d = 39 + (5 - 1) \cdot 130 = 559$

$a_{6} = a_{1} + (n - 1) \cdot d = 39 + (6 - 1) \cdot 130 = 689$

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Why learn this

When will the next bus arrive? How many people can fit inside a stadium? How much money will I earn this year? All these questions can be answered by learning how arithmetic sequences work. The progression of time, triangular patterns (bowling pins, for example), and increases or decreases in quantity can all be expressed as arithmetic sequences.

Terms and topics

Arithmetic sequences