Question

Find an equation for the nth term of a geometric sequence where the second and fifth terms are -2 and 16, respectively.

a. an = 1 • (-2)n - 1
b. an = 1 • 2n
c. an = 1 • (-2)n + 1
d. an = 1 • 2n - 1

Answer 1

know the formula for n'th term of geometric sequence ?

Answer 2

Is it: a(n) = a*r^n-1 ?

Answer 3

yes!
\(\Large a_n = a_1 r^{n-1}\)

Answer 4

Okay. So where do I go from there?

Answer 5

so for 2nd term, put n =2
and a2 = -2
what do u get ?

Answer 6

\[a _{n} = -2r ^{2-1}\]
Is that what you mean?

Answer 7

hi hi hi

Answer 8

but we don't know a_1
-2 is actually a_2 , not a_1

Answer 9

\(\large a_2 = a_1 r^{2-1} \\ -2 =a_1 r \)
got this ?

Answer 10

So would I be using a different formula then?

Answer 11

Oh okay.

Answer 12

Yeah, I got it. One sec...

Answer 13

Okay so now that we have that formula, what do I do? How do I know what r is?

Answer 14

we also have 5th term!

put n=5
and a5 = 16
in the same formula

Answer 15

\[16 = a1^{r4}\]

Answer 16

Is that wrong or completely wrong?

Answer 17

\(\Large 16 = a_1 r^4\)
thats correct!
now we need to find 'r'
know how to find r from these 2 equations ???
hint : divide

Answer 18

16/4 = 4
r = 4

Answer 19

Is that right or did I skip steps?

Answer 20

-2 = a1 r
16 = a1 r^4

dividing
-2/16 = r/r^4

got this ?

Answer 21

note we divided , so that a1 gets cancelled and we could get r

Answer 22

so r = 8?

Answer 23

or is it r = -2

Answer 24

r^3 = -8
so, yeah, r= -2 :)

Answer 25

Okay :) So what's next?

Answer 26

-2 = a_1 r
you know r
find a1

Answer 27

How do I apply all of that into this?:

a. an = 1 • (-2)n - 1
b. an = 1 • 2n
c. an = 1 • (-2)n + 1
d. an = 1 • 2n - 1

Answer 28

coming to it :)

Answer 29

Okay.

Answer 30

-2 = a1 r
and r = - 2
so does a1 = 1? or 0?

Answer 31

-2 = a1 * (-2)
a1 = -2/-2 = 1
:)
now plug the values in general formula!
\(\Large a_n = a_1 r^{n-1}\)

Answer 32

\[a _{n} = 1(-2) a _{n} = -2\]

Answer 33

an = -2 correct?

Answer 34

keep an as an
\(\Large a_n = 1 \times (-2)^{n-1}\)
is than an option ?

Answer 35

Oh okay. And yes, it is. Thanks so much for helping me!

Answer 36

welcome ^_^

Answer 37

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