How do you find 37th term of the sequence {4, 7, 10, ....}?

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How do you find 37th term of the sequence {4, 7, 10, ....}?

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Answer 1 · 2018-08-02T09:19:33

$112$ $112$ is the 37th term

Explanation:

This is an arithmetic sequence

So, $T_{n} = a + (n - 1) d$ $T_n=a+(n-1)d$
where $a$ $a$ is your first term, $n$ $n$ is your nth term and $d$ $d$ is the difference between 2 adjacent terms

Looking at the sequence,
$a = 4$ $a=4$
$d = 3$ $d=3$

Since you want to find the 37th term, then $n = 37$ $n=37$

$T_{n} = a + (n - 1) d$ $T_n=a+(n-1)d$
$T_{37} = 4 + (37 - 1) 3$ $T_37=4+(37-1)3$
$T_{37} = 4 + 36 \times 3$ $T_37=4+36times3$
$T_{37} = 112$ $T_37=112$

Answer 2 · 2018-08-02T09:22:29

$a_{37} = 112$ $a_37=112$

Explanation:

$these are the terms of an arithmetic sequence$ $"these are the terms of an arithmetic sequence"$

$the n th term of an arithmetic sequence is$ $"the n th term of an arithmetic sequence is"$

$∙ x a_{n} = a_{1} + (n - 1) d$ $•color(white)(x)a_n=a_1+(n-1)d$

$where a_{1} is the first term and d the common difference$ $"where "a_1" is the first term and d the common difference"$

$d = 7 - 4 = 10 - 7 = 3 and a_{1} = 4$ $d=7-4=10-7=3" and "a_1=4$

$a_{37} = 4 + (36 \times 3) = 4 + 108 = 112$ $a_(37)=4+(36xx3)=4+108=112$

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How do you find 37th term of the sequence {4, 7, 10, ....}?

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