anonymous
  • anonymous
FAN/MEDAL What is the sum of an 8-term geometric series if the first term is -11, the last term is 859,375, and the common ratio is -5? -143,231 -36,047 144,177 716,144
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
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anonymous
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anonymous
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anonymous
  • anonymous
amoodarya
  • amoodarya
\[S_8?\\a_1=-11\\a_8=859,375\] \[s_n=a_1+a_1q+a_1q^2+...+a_1q^{n-1}=\frac{a_1(1-q^n)}{1-q}\\s_8=a_1+a_1q+a_1q^2+...+a_1q^{7}=\frac{a_1(1-q^8)}{1-q}\]
Michele_Laino
  • Michele_Laino
hint: the sum of the first 8 terms of your geometric sequence is given by the subsequent formula: \[\Large S = {a_1}\frac{{{q^8} - 1}}{{q - 1}}\]
Michele_Laino
  • Michele_Laino
where q=-5 and a_1= -11
amoodarya
  • amoodarya
you can write it as \[a_8=a_1q^7=-11(q^7)=859,375\\q=-5\] plug into formula
anonymous
  • anonymous
Michele_Laino
  • Michele_Laino
substituting your data into the formula for S, we get: \[\Large S = {a_1}\frac{{{q^8} - 1}}{{q - 1}} = \left( { - 11} \right) \times \frac{{{{\left( { - 5} \right)}^8} - 1}}{{\left( { - 5 - 1} \right)}} = ...?\] please continue
anonymous
  • anonymous
-390625-1 (-11) -390626 ---------- x ---- = -------- = -65104.33333 x -11 = -716147.6667 -6 1 6
anonymous
  • anonymous
I got this wrong, right?
Michele_Laino
  • Michele_Laino
please keep in mind that: \[\Large {\left( { - 5} \right)^8} = 390625\]
anonymous
  • anonymous
But then I added to there is a 1- in front
anonymous
  • anonymous
3906254 (-11) 39062 ---------- x ---- = -------- = -65104 x -11 = -716144 -6 1 - 6
anonymous
  • anonymous
I got it is that right?
Michele_Laino
  • Michele_Laino
\[{\left( { - 5} \right)^8} = {\left( { - 1 \times 5} \right)^8} = {\left( { - 1} \right)^8}{\left( 5 \right)^8} = + {5^8} = 390625\] hint: \[S = {a_1}\frac{{{q^8} - 1}}{{q - 1}} = \left( { - 11} \right) \times \frac{{{{\left( { - 5} \right)}^8} - 1}}{{\left( { - 5 - 1} \right)}} = \left( { - 11} \right) \times \frac{{390625 - 1}}{{\left( { - 5 - 1} \right)}} = ...?\]
anonymous
  • anonymous
But it's positive
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
Yes its positive negative times negative
anonymous
  • anonymous
Thanks! Can you help me with my next question
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
What is the 7th term of the geometric sequence where a1 = 1,024 and a4 = -16? 1 -0.25 -1 0.25
Michele_Laino
  • Michele_Laino
we have the subsequent formulas: \[\begin{gathered} {a_4} = {a_1}{q^3} \hfill \\ {a_7} = {a_1}{q^7} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
so we can re-write the first formula as below: \[ \Large - 16 = 1024 \times {q^3}\] please solve for q
anonymous
  • anonymous
-16/1024 = 1024 x q^3/1024 -.015625=q^3 (3)sqrt(-.015625)=(3)sqrt(q^3) q=.53860
Michele_Laino
  • Michele_Laino
the computation -.015625=q^3 is right, so we have: \[\large q = \sqrt[3]{{ - 0.015625}} = ...?\]
Michele_Laino
  • Michele_Laino
I got: q=-1/4 am I right?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Oh I got it now, I forgot to put the negative on the calculator
Michele_Laino
  • Michele_Laino
ok! Now we have: \[\Large {a_7} = {a_1}{q^7} = 1024 \times {\left( { - \frac{1}{4}} \right)^7} = ...?\]
anonymous
  • anonymous
0.25?
Michele_Laino
  • Michele_Laino
sorry, I got another result
anonymous
  • anonymous
I got now -.0625
Michele_Laino
  • Michele_Laino
yes! \[\Large {a_7} = - 0.0625 = - \frac{1}{{16}}\]
anonymous
  • anonymous
Thanks!
Michele_Laino
  • Michele_Laino
:)

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