Question

Assume that the length of Figure A is ½ inch and the length of Figure B is 1 inch. Use a ratio to compare the length of Figure A to the length of Figure B. What is this ratio?

Answer 1

@Michele_Laino

Answer 2

please where is the drawing of figures A, and B?

Answer 3

sorry

Answer 4

the ratio, is:
\[\Large r = \frac{{figure\;A}}{{figure\;B}} = \frac{{1/2}}{1} = \frac{1}{2}\]

now, the length of the base of figure B is 10 cm, so the length of the base of figure A, will be:
\[\Large 10r = \frac{10}{2} = ...?\]

Answer 5

ok so the ratio will be 1/2
now I have two more parts to it
If the same ratio is used to enlarge Figure B and create a Figure C, what would the length of the new figure be?

Write the geometric sequence using the three lengths.
What is a1?
What is the common ratio r?
Write an explicit formula for an using the formula for geometric sequences an = a1*(r)n – 1.

Answer 6

@Michele_Laino

Answer 7

i think 20

Answer 8

the length of the new figure C is 2 times the length of figure B
\[\Large r = \frac{{figure\;B}}{{figure\;C}} = \frac{{1/2}}{1} = \frac{1}{2}\]

Answer 9

so, we get a geometric sequence whose common ratio is \(2\)

Answer 10

But we have to use the measurements they give us which is 1/2 and 1 in which cas figure c would be 2

Answer 11

if the first term is \(1/2\), then we can write:
\[\Large {a_1} = \frac{1}{2},\quad {a_2} = 1,\quad {a_3} = 2\]

Answer 12

ok So a1 is 1/2

Answer 13

yes!

Answer 14

the common ratio is what

Answer 15

1/2?

Answer 16

it is \(2\), since:
\[\Large common\;ratio = \frac{{{a_3}}}{{{a_2}}} = \frac{{{a_2}}}{{{a_1}}}\]

Answer 17

ok Now to the equation part

Answer 18

an= 1/2 (2)^n-1?

Answer 19

@Michele_Laino

Answer 20

we have the subsequent formulas:
\[\Large \begin{gathered}
{a_1} = \frac{1}{2} \hfill \\
\quad \hfill \\
{a_2} = 2{a_1} \hfill \\
\hfill \\
{a_3} = 2{a_2} = 4{a_1} = {2^2}{a_1} = {2^{3 - 1}}{a_1} \hfill \\
\end{gathered} \]

Answer 21

if we continue it looks that the the formula for the n-th term is:
\[\Large {a_n} = {2^{n - 1}}{a_1}\]

Answer 22

so the whole thing would be part of my answer?

Answer 23

only the last formula:
\[\Large {a_n} = {2^{n - 1}}{a_1}\]

Answer 24

why would a1 be at the end wopuldnt it be 1/2*2 ^n-1

Answer 25

the n-th term is:
\[\Large {a_n} = {2^{n - 1}}\left( {\frac{1}{2}} \right) = {2^{n - 2}}\]

Answer 26

what is the n-th term

Answer 27

it is a generic term, for example, if \(n=5\), then we have:
\[{a_5} = {2^{5 - 1}}\left( {\frac{1}{2}} \right) = {2^{5 - 2}} = {2^3} = 8\]

Answer 28

an=2^n−1 a1

Answer 29

so this is my answer?

Answer 30

Yes! It is the same formula that I wrote

Answer 31

correct i was just confused so i wanted to make sure

Answer 32

ok! :)