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belka [17]
3 years ago
6

Find the first six terms of the sequence.

Mathematics
1 answer:
Levart [38]3 years ago
3 0
A(n) = a1 * r^(n-1)
a1 = first term = -8
r = common ratio = 5

a(2) = -8 * 5^(2-1)
a(2) = -8 * 5^1
a(2) = -40

a(3) = -8 * 5^(3 - 1)
a(3) = -8 * 5^2
a(3) = -8 * 25
a(3) = -200

a(4) = -8 * 5^(4 - 1)
a(4) = -8 * 5^3
a(4) = -8 * 125
a(4) = -1000

a(5) = -8 * 5^(5 - 1)
a(5) = -8 * 5^4
a(5) = -8 * 625
a(5) = -5000

a(6) = -8 * 5^(6-1)
a(6) = -8 * 5^5
a(6) = -8 * 3125
a(6) = -25,000

the first 6 terms are : -8, -40, -200, -1000,  -5000, -25,000
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Answer:

1

Step-by-step explanation:

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).

Thus:

\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) }

Let's do the first part first: (Recall how to divide fractions)

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

For the second term:

\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)

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

Note the numerator; it can be derived from the Pythagorean Identity:

sin^2(x)+cos^2(x)=1; cos^2(x)=1-sin^2(x)

Thus, we can substitute the numerator:

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

Everything simplifies to 1.

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