Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

Write a formula for the general term (the nth term) of the geometric sequence. 1/2 -1/10 1/50 -1/250

OpenStudy (anonymous):

@phi @welshfella

OpenStudy (anonymous):

A. an = 1/2(n-1)3/5 B. an = 1/2 - 1/5(n-1) C. an = (1/2)(-1/5)(n-1) D. an = (1/5)(-1/2)(n-1)

OpenStudy (phi):

any idea how to "get" from 1/2 to -1/10 ? what do you multiply 1/2 by to get -1/10 ?

OpenStudy (anonymous):

5 maybe

OpenStudy (phi):

right track. but you can do better \[ \frac{1}{2} \cdot x =- \frac{1}{10} \]

OpenStudy (phi):

if you multiply by 5 you get \[ \frac{5}{2} \] not what we want

OpenStudy (anonymous):

1/5

OpenStudy (phi):

if you can't see it, try multiplying both sides by 2 and "solve for x" 1/5 is pretty good, but not quite \[ \frac{1}{2} \cdot \frac{1}{5}= \frac{1}{10} \] but we want \[ - \frac{1}{10} \]

OpenStudy (anonymous):

so -1/5

OpenStudy (phi):

ok, and to make sure take the 2nd term in the series -1/10 and multiply by -1/5 what do we get ?

OpenStudy (anonymous):

1/15

OpenStudy (anonymous):

i think

OpenStudy (phi):

multiply top times top and bottom times bottom

OpenStudy (anonymous):

ok so 1/50

OpenStudy (phi):

and if we multiply 1/50 by -1/5 we get -1/250 notice we are getting the original series 1/2 -1/10 1/50 -1/250

OpenStudy (anonymous):

yes

OpenStudy (phi):

here is what we know \[ \frac{1}{2} \\ \frac{1}{2}\left(-\frac{1}{5}\right)^1 \\ \frac{1}{2}\left(-\frac{1}{5}\right)^2 \] and so on

OpenStudy (anonymous):

yes

OpenStudy (phi):

if we label the first term as "term number 1" and the 2nd 2 and then 3 etc notice each term has 1/2 times (-1/5) to a power one less than the term's number

OpenStudy (anonymous):

yes that makes sence

OpenStudy (phi):

only one of the choices is "short-hand" for that

OpenStudy (anonymous):

c

OpenStudy (phi):

yes

OpenStudy (anonymous):

yeah thnx again freind

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!