anonymous
  • anonymous
given an infinite geometric progression with the product of the first 3 consecutive terms is 216 and the sum of the numbers is 26,find the sum of all the terms
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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terenzreignz
  • terenzreignz
Pretty :D Let's have the first three terms as \(\large a\), \(\large ar\) and \(\large ar^2\) shall we? ^_^
terenzreignz
  • terenzreignz
@Jonask
anonymous
  • anonymous
yes thats okay

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terenzreignz
  • terenzreignz
Well then, their product, which is \(\large a^3 r^3\) is equal to 216 Which can only mean that \(\large ar = 6\)
anonymous
  • anonymous
let the terms be a/r ,a, ar such that a/r * a * ar= 216 which can easily give u a =6 again given 6/r + 6+6r=26 so to get r solve 6/r +6r =20 ...
terenzreignz
  • terenzreignz
And then their sum... \[\Large a+ar + ar^2=26\]\[\Large a+ar+ar\cdot r=26\]\[\Large a+ 6+ 6r = 26\]etc...
anonymous
  • anonymous
yeah nice i did it quite differently \[a_1a_2a_3=216\\since \frac{a_2}{a_1}=\frac{a_3}{a_2}\implies a_2^2=a_1 a_3\\a_1a_2a_3=a_2a_2^2=a_2^3=216\\a_2=6\\a_1+a_2+a_3=26\implies a_1+a_3=20,a_1a_3=6^2=36\\hence\\a_1=8....a_3=2\\a_1,a_2,a_3=8,6,2\] \[S_\infty=\frac{8}{1-3/4}\]
terenzreignz
  • terenzreignz
So... was there even a problem? :/
anonymous
  • anonymous
i dont know
anonymous
  • anonymous
hence Given that the product of the first n terms of a geometric sequence is P ,find \[\large\prod_{k=1}^na_{3k-1}\]

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