Question

Find the general term an for the geometric sequence
a1= -4 and a2 = 20

Answer 1

\[a _{n} a _{1} = -4 a _{2} = 20 \]

Answer 2

@satellite73 help?

Answer 3

\[a_1~r = a_2\]what is r?

Answer 4

he didn't give me R he only gave me a1 and a2

Answer 5

.... then you might want to plug that a1 and a2 into the formula and SOLVE for r

Answer 6

what is that formula?

Answer 7

the a1+ (n-1)d

Answer 8

no, we do not want to find an arithmetic sequence with a common difference (d)

we want to find an geometric sequence with a ratio (r)

a geometric sequence is the result of multiplying the same value (r) to each new an in order to determine the next value.

therefore, you are given 2 values, a1 and a2. By applying the definition of a geometric sequence, the value of a2 must be a1 times r

Answer 9

therefore:

\[a_1~r=a_2\]

what is the common ratio we want to determine? what is the value of "r" ?

Answer 10

well to get a2 to be 20, you can multiply -4 by -5

Answer 11

correct :)

so we have 2 ways we can express the \(a_n\) term. recurrsively, and also explicitly

which way do you need to express it?

Answer 12

he doesn't state in which way to express it

Answer 13

recurrsively is defining the n+1 in terms of n\[a_{n+1}=a_{n}~r~;~a_1 = k\]

the other way ...

Answer 14

explicitly stated:\[a_n=a_1 ~r^{n-1}\]

Answer 15

Let me look and see how they are showing this in the chapter, it's only suppose to be an algebra class

Answer 16

ok, im assuming they want the explicit then :)

Answer 17

okay, your way better at this then me. do we assume n = 1 ?

Answer 18

n is the value of the nth term of the sequence that we would want to fine, its best to leave it as the variable in the equation so that we can fill it in as needed

Answer 19

okay leaving the equation to look like a n = -4(-5) n-1

Answer 20

spose we wanted to know the 156th term of the sequence

the recurrsion equation would take some time to get all the way up to 156

the explicit equation allows us a more convient method:\[a_{156}=a_1~r^{156-1}\]

Answer 21

yes:\[a_n=-4~(-5)^{n-1}\]

Answer 22

so for the general term is an= -4(-5) n-1

Answer 23

yes, make sure you exponent that (n-1) tho.

Answer 24

have time for one more?

Answer 25

maybe, hard to tell the future :)

Answer 26

Write the terms of their series and find their sums \[\sum_{k=1}^{4} (k+5)^{2} \]

Answer 27

well, id write out (k+5)^2 .... enough times to calculate k=1,2,3,4 ; so four times

Answer 28

then for each time youve written it, fill in a new k value and evaluate the setup

Answer 29

1: (k+5)^2
2: (k+5)^2
3: (k+5)^2
4: (k+5)^2



1: (1+5)^2
2: (2+5)^2
3: (3+5)^2
4: (4+5)^2



1: (6)^2
2: (7)^2
3: (8)^2
4: (9)^2

Answer 30

on our homework we had the same equation but the top was 4 and k=3, he had us figuring it out by k= 3^2-3= 9-3 = 6
4^2 -4 = 16-4 = 12
12+6 = 18

Answer 31

ok, then the question is asking you to find partial sums, or a running cumulation of the sums of the sequnce that is generated

Answer 32

I had to find the series and it's sum of the equation.

Answer 33

a1: (6)^2
a2: (7)^2
a3: (8)^2
a4: (9)^2


S1 = a1
S2 = a1 + a2
S3 = a1 + a2 + a3
S4 = a1 + a2 + a3 + a4

Answer 34

so the series would be 36,49, 64, and 81

Answer 35

the "sequence" is 36,49,64, and 81

the "series" is: 36 + 49 + 64 + 81

Answer 36

so 230

Answer 37

sounds about right

Answer 38

okay so 4 ∑k=1(k+5)2 how do we solve for that?

Answer 39

you just did, that is what we just went thru

Answer 40

4 ∑k=1(k+5)2 means: (1+5)^2 + (2+5)^2 + (3+5)^2 + (4+5)^2

its just more concise to write it as: 4 ∑k=1(k+5)2 to express the same thing

Answer 41

ooh okay, figured that out now. the way he set up the question threw me off. Okay so the terms of the sequence is 36,49,64,81

Answer 42

correct

Answer 43

okay, I am sorry I am not trying to make this harder on you, I am just so lost in this class.

Answer 44

if its new material, then there is a bit of a learning curve :) im just used to the notations better

Answer 45

high school algebra sure has changed since I went. I never had to do things like this so it is a HUGE learning curve.

Answer 46

i was 20 years out of high school and went to college to get a masters in math. my first class i found out that i had retained half of my school math, forgot about 1/4 of it; and there was 1/4 i had never even come across

Answer 47

it's been about 5 year since high school, getting my degree in the medical field. Thank you so much for your help and taking the time . I have two more problem to see if anyone can help with =)

Answer 48

good luck :) ill be around ....

Answer 49

thank you.