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2 years ago

14

Indicate a general rule for the n th term of this sequence. 6a, 3a, 0, -3a, -6a, . . .

1 answer:

GrogVix [38]2 years ago

3 0

Hello :
(-6a)-(-3a) =(-3a)-(0) = (0)-(3a) = (3a)-(6a) = -3a (arthmetic<span> sequence, the common - diff is : d= -3a the first term is : U1 = 6a
</span><span>a general rule for the n th term is : Un =U1+(n-1)d
Un = 6a +(n-1)(-3a) =
Un = 6a -3an+3a
</span>Un = -3an +9a

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csec%5Cleft%28x%5Cright%29%7D%7B%5Ccos%5Cleft%28x%5Cright%29%7D-%5Cfrac%7B%5Csin%5

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First, convert all the secants and cosecants to cosine and sine, respectively. Recall that $csc(x)=1/sin(x)$ and $sec(x)=1/cos(x)$ .

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$\frac{sec(x)}{cos(x)} -\frac{sin(x)}{csc(x)cos^2(x)}$

$=\frac{\frac{1}{cos(x)} }{cos(x)} -\frac{sin(x)}{\frac{1}{sin(x)}cos^2(x) }$

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$\frac{\frac{1}{cos(x)} }{cos(x)}=\frac{1}{cos(x)} \cdot \frac{1}{cos(x)}=\frac{1}{cos^2(x)}$

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$\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}$

So, all together: (same denominator; combine terms)

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Everything simplifies to 1.

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