anonymous
  • anonymous
I will fan and Medal!!!! Please help if you can! The sum of first three terms of a finite geometric series is -7/10 and their product is -1/125. [Hint: Use a/r, a, and ar to represent the first three terms, respectively.] The three numbers are _____, _____, and _____.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amoodarya
  • amoodarya
\[a+aq+aq^2=-\frac{7}{10}\\a*(aq)*(aq^2)=-\frac{1}{125}\] is it ?
amoodarya
  • amoodarya
a little trick , if we use it is simpler \[a_1=\frac{a}{q} \\a_2=a\\a_3=aq\\now\\\frac{a}{q}+a+aq=-\frac{7}{10}\\\frac{a}{q}*a*aq=-\frac{1}{125}\rightarrow a^3=\frac{-1}{125} \rightarrow a=\frac{-1}{5}\\\] now put a=-1/5 to other equation and find q
amoodarya
  • amoodarya
can you go on ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amoodarya
  • amoodarya
are you there ?
amoodarya
  • amoodarya
\[a=\frac{-1}{5}\\\frac{a}{q}+a+aq=\frac{-7}{10}\\\frac{\frac{-1}{5}}{q}+\frac{-1}{5}+\frac{-1}{5}q=\frac{-7}{10}\\\frac{1}{q}+1+q=3.5\]
anonymous
  • anonymous
thanks alot @amoodarya
amoodarya
  • amoodarya
your welcome

Looking for something else?

Not the answer you are looking for? Search for more explanations.