3 years ago

What is the formula for the following arithmetic sequence? 12, 6, 0, -6, ...

Mathematics

2 answers:

bixtya [17]3 years ago

6 0

Answer:

an= 12 + (-6)(n-1)

Step-by-step explanation:

you should find two values: a1 and d

a1= first number =12

d= a number of the sequence mines the number of the sequence before the one you chose = 0-6 = -6

a1+(n-1)d= 12-6(n-1)

mina [271]3 years ago

3 0

Answer:

12 + (-6)(n-1)

Hope it helped <3

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If A+B+C=<img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\pi" align="absmiddle" class="latex-formula"

seraphim [82]

Answer:

$a + b + c = \pi \\ = > c= \pi - a - b \\ < = > \tan(c) = \tan(\pi - a - b) = -\tan(a + b)$

Step-by-step explanation:

we have:

$\tan(a) + \tan(b) + \tan(c) \\ = \tan(a) + \tan(b) - \tan(a + b) \\ = \tan( a) + \tan(b) - \frac{ \tan(a) + \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{ ( \tan(a) + \tan(b) ) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (1)$

we also have:

$\tan(a) \tan(b) \tan(c) \\ = - \tan(a) \tan(b) \tan(a + b) \\ = \frac{ -(\tan( a ) + \tan(b) ) \tan(a) \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{( \tan(a) + \tan(b)) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (2)$