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klemol [59]
2 years ago
14

If the 11th term of a geometric sequence is 32 times larger than the 6th term, then

Mathematics
1 answer:
mojhsa [17]2 years ago
4 0

Answer:

r = 2

Step-by-step explanation:

nth term of a geometric sequence = ar^{n - 1}.

Where,

a = first term, r = common ratio,

We are given that, the 11th term is 32 times larger than the 6th term of the geometric sequence. Thus:

the 6th term can be expressed as ar^{n - 1} = ar^{6 - 1}

a_6 = ar^{5}

The 11th term is expressed as a_11 = ar^{10}

Since the 11th term is 32 times larger than the 6th term, therefore, the following equation can be derived to find the common ratio, r, of the sequence:

a_11 = 32*a_6

a_11 = ar^{10}

a_6 = ar^{5}, therefore:

ar^{10} = 32*ar^{5}

Divide both sides by ar⁵

\frac{ar^{10}}{ar^5} = \frac{32*ar^{5}}{ar^5}

r^5 = 32

r^5 = 2^5

r = 2

Common ratio = 2

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Given an angle. construct an angle congruent to the given angle.

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Construct: An angle congruent to angle ABC

Procedure:

1. Draw a ray. Label it ray RY.

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4. Using S as center and a radius equal to DE, draw an arc that intersects arc XS at a point Q.

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Justification (for congruence): If you draw line segment DE and line segment QS, triangle DBE is congruent to triangle QRS (SSS postulate) Then angle QRS is congruent to angle ABC.

You can probably also Google videos if it's hard to imagine this. Sorry, construction is super hard to describe.

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