Question

Given a geometric sequence in the table below, create the explicit formula and list any restrictions to the domain.


n an
1 −4
2 20
3 −100
an = −5(−4)n − 1 where n ≥ 1

an = −4(−5)n − 1 where n ≥ 1

an = −4(5)n − 1 where n ≥ −4

an = 5(−4)n − 1 where n ≥ −4

@D3xt3R

Answer 1

We've got a sequence\[a_1=-4\\a_2=20\\a_3=-100\]What do we have in common?!\[a_1=-4\\a_2=(-4)*(-5)=20\\a_3=(-4)*(-5)*(-5)=(-4)*(-5)^2=-100\]

Answer 2

? @D3xt3R

Answer 3

You have to finish the exercise, how can you make a general term?!

Answer 4

idk, i didnt learn this. im taking the class over online

Answer 5

Think with me.\[a_1=-4\]\[a_2=(-4)*(-5)\]\[a_3=(-4)*(-5)^2\]Do you understand?!

Answer 6

no

Answer 7

\[a_1=-4\\a_2=20\\a_3=-100\]Do you understand it?!

Answer 8

ok

Answer 9

What do you think, if I say the general term is\[a_n=a_1*(-5)^{n-1}\]??? Do you understant what I do?!

Answer 10

@cwrw238

Answer 11

the terms in a geometric series are produced by multiplying b y a constant this is called the common ratio
a simple example would be

1 , 2 , 4 , 8

where each term is multiplied by 2 to get the next one

1*2 = 2
2 * 2 = 4
4 * 2 = 8 etc

here the common ratio is 2

common ratio is usually denoted by r and first term as a

so we can write the sequence as

a , ar, ar^2 , ar^3

note that the power on r is going up 1 at a time and the 3rd term has power 2, the fourth term has power 3 , and so on.

so we can write the general nth term as a r^(n-1)

Answer 12

another convention is to write the first , second. third terms as a1, a2 , a3 etc

Answer 13

ok

Answer 14

notice also that you can find the common ratio r by dividing the first term by the first, or third term by the second , 4th by the third etc

because ar/a = r and ar^2 / ar = r etc

so to find r in the sequence in the question
we can divide 2nd by the first - that is 20 / -4 = -5
-100 / 50 = -5

so common ratio r = -5

Answer 15

so to find an - that id the nth term of the series in the wuestion
we have the general formula

an = ar^(n-1)

now a = -4 and r = -5

so an = -4(-5)^(n-1)

now n being the sequance number of the term must be 1 or greater

so we write n>+1

Answer 16

n >= 1

so the 2nd option is the correct one.

Answer 17

Formula to commit to memory for a GS is

nth term = ar^(n-1)

or if you like nth term = a1r^(n-1)

a or a1 is the first term

Answer 18

thank you