// Numbas version: exam_results_page_options {"name": "Finding the formula for the $n^{\\text{th}}$ term of linear sequences", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Finding the formula for the $n^{\\text{th}}$ term of linear sequences", "type": "question", "statement": "

A linear sequence is a series of numbers that either increases or decreases by a constant amount at each step.

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Find formulas for the $n^{\\text{th}}$ term for each of the following linear sequences, where the values for $n=1\\text{,}2\\text{,}3$ are given:

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$\\var{m[1]*2}, \\var{m[1]*3}, \\var{m[1]*4}, \\ldots$

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$n^\\text{th}$ term = [[0]]

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The formula for the $n^{\\text{th}}$ term of an arithmetic sequence is

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\$a_n = a_1 + (n-1)d, \$

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where

\n
\n
• $a_n$ is the $n^\\text{th}$ term;
• \n
• $a_1$ is the first term in the sequence;
• \n
• $d$ is the common difference between consecutive terms.
• \n
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For this arithmetic sequence, what is $a_1$?

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What is $d$?

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$\\var{m[2]*8+2}, \\var{m[2]*7+2}, \\var{m[2]*6+2}...$

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$n^\\text{th}$ term = [[0]]

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The formula for the $n^{\\text{th}}$ term of an arithmetic sequence is

\n

\$a_n = a_1 + (n-1)d, \$

\n

where

\n
\n
• $a_n$ is the $n^\\text{th}$ term;
• \n
• $a_1$ is the first term in the sequence;
• \n
• $d$ is the common difference between consecutive terms.
• \n
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For this arithmetic sequence, what is $a_1$?

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What is $d$?

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Given the first three terms of a sequence, give a formula for the $n^\\text{th}$ term.

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In the first sequence, $d$ is positive. In the second sequence, $d$ is negative.

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Both of these sequences are linear, or arithmetic, sequences. To find formulas for these sequences we need to identify their first terms and common differences.

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#### a)

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The formula for the $n^\\text{th}$ term of an arithmetic sequence is

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\$a_n = a_1 + (n-1)d \\text{.} \$

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$a_1$ is the first term and $d$ the common difference between consecutive terms, which we need to identify.

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We can find these by drawing up a table of $a_n$ against $n$, and the differences between consecutive terms.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $n$ 1 2 3 $a_n$ $\\pmb{\\var{m[1]*2}}$ $\\var{m[1]*3}$ $\\var{m[1]*4}$ First differences $\\pmb{\\var{m[1]}}$ $\\pmb{\\var{m[1]}}$
\n

The first term and common difference have been highlighted in bold; we can use these to write the formula for the sequence.

\n

\\begin{align}
a_n &= a_1+(n-1)d \\\\
&= \\var{m[1]*2}+(n-1)\\times\\var{m[1]} \\\\
&= \\var{m[1]}n + \\var{m[1]}\\text{.}
\\end{align}

\n

#### b)

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Similar to the part a), we can identify $a_1$ and $d$ for this sequence by drawing a table of $a_n$ against $n$.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $n$ 1 2 3 $a_n$ $\\pmb{\\var{m[2]*8+2}}$ $\\simplify{{m[2]}*7+2}$ $\\simplify{{m[2]}*6+2}$ First differences $\\pmb{\\var{-m[2]}}$ $\\pmb{\\var{-m[2]}}$
\n

The first term and common difference have been highlighted in bold; we can use these to form the formula for the sequence.

\n

\\begin{align}
a_n &=a_1+(n-1)d \\\\
&=\\var{m[2]*8+2}+(n-1)\\times\\var{-m[2]} \\\\
&=-\\var{m[2]}n + \\var{m[2]*9+2}\\text{.}
\\end{align}

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