Given the nth term for each arithmetic sequence, how to find the indicated term here? (1) $a_{n} = 6 n + 1; a_{12}$ $a_n=6n+1; a_12$ (2) $a_{n} = 3 n - 5; a_{20}$ $a_n=3n-5; a_20$

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Given the nth term for each arithmetic sequence, how to find the indicated term here? (1) $a_{n} = 6 n + 1; a_{12}$ $a_n=6n+1; a_12$ (2) $a_{n} = 3 n - 5; a_{20}$ $a_n=3n-5; a_20$

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Answer 1 · 2018-06-27T08:25:39

1) 73
2) 55

Explanation:

1) $a_{n} = 6 n + 1$ $a_(color(red)(n)) =6color(red)(n)+1$

For $a_{12}$ $a_12$ where ever there is an $n$ $color(red)(n)$ write 12 giving:

$a_{12} = 6 (12) + 1 d \to d a_{12} = [6 \times 12] + 1 = 73$ $a_12=6(12)+1 color(white)("d") ->color(white)("d")a_12=[6xx12]+1=73$

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2) $a_{n} = 3 n - 5$ $a_color(red)(n)=3color(red)(n)-5$

For $a_{20}$ $a_color(red)(20)$ where ever there is an $n$ $color(red)(n)$ write 20 giving:

$a_{20} = 3 (20) - 5 d \to d a_{20} = [3 \times 20] - 5 = 55$ $a_color(red)(20)=3(color(red)(20))-5color(white)("d")->color(white)("d")a_20=[3xx20]-5=55$

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Given the nth term for each arithmetic sequence, how to find the indicated term here? (1) $a_{n} = 6 n + 1; a_{12}$ $a_n=6n+1; a_12$ (2) $a_{n} = 3 n - 5; a_{20}$ $a_n=3n-5; a_20$

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Given the nth term for each arithmetic sequence, how to find the indicated term here? (1) an=6n+1;a12a_n=6n+1; a_12 (2) an=3n−5;a20a_n=3n-5; a_20

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Given the nth term for each arithmetic sequence, how to find the indicated term here? (1) $a_{n} = 6 n + 1; a_{12}$ $a_n=6n+1; a_12$ (2) $a_{n} = 3 n - 5; a_{20}$ $a_n=3n-5; a_20$