HOW DO I FIND THE AVERAGE COMMON RATIO OF A GEOMETRIC SERIES.
if its a geometric sequence we have the setup: \[g,~gr,~gr^2,~gr^3,...\] \[\frac{gr^k}{gr^{k-1}}=r\]
in other words, just take some term, and divide it by the term before it ... 1,3,9,18, 3/1 = 3 9/3 = 3 18/9 = 3 etc ....
hmmm, that seems a bit more complicated :)
yeah. :/ @amistre64
a ratio a:b can be considered a fraction a/b an average is a sum divided by how many there are .... one idea may be that you can add up 5 ratio amounts and divide by 5 ... naybe
or, if we are trying to get a geometric sequence that starts at 3 and ends at 1.2 in 5 terms ...
yes, i have 5 terms, the first number is 3, and the last number is 1.2. i need to know what the average common ratio of that is.. i dont understand how to get it.
\[g_n = g_{n-1}~r\] \[g_n = g_{0}~r^{n-1}\] \[1.2 = 3~r^{4}\] divide and 4rt
or work some logs
@chell: Please post your own question separately. Type it into the "Ask a question" box.
@amistre64 okay. so 1.2 = 3 r^4 and solve that? and that's my common ratio?
if we are wanting to form a geometric sequence that starts at 3, and the 5th term is 1.2 .... then yes
okay so that's 0.795 that's my common ratio?
sounds good to me .... assumptions in tow
to keep accuracy, 4rt(.4) might be warranted as an exact result
@amistre64 can you help me with this part???
What is the height of each ball on the fifth bounce (i.e., Height 6)? Use the geometric sequence formula, an = a1rn – 1 and show your work. a1 = r^n-1
well, we know r, and the height for 5 ... so multiply r by H5 to get H6
what is 1.2, times 4rt(.4)?
1.272
1.2 * .795 is not more than 1.2 ....
79% of 1.2 is not greater than 100% of 1.2 :)
oh its .954 what does that have to do with this though?
well, 3*r^4 = 1.2 for height 5. what is height 6? tack on another r :/ ... 3*r^5
yes. @amistre64 that makes sense.
wow, i just got a notice that someone needs to learn something by today ... procrastination is such a blessing :)
@amistre64 thank you for your help. i really appreciate it. (:
good luck ;)
Join our real-time social learning platform and learn together with your friends!