Question

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

Answer 1

Pretty :D
Let's have the first three terms as \(\large a\), \(\large ar\) and \(\large ar^2\) shall we?

^_^

Answer 2

@Jonask

Answer 3

yes thats okay

Answer 4

Well then, their product, which is \(\large a^3 r^3\) is equal to 216
Which can only mean that \(\large ar = 6\)

Answer 5

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 ...

Answer 6

And then their sum...
\[\Large a+ar + ar^2=26\]\[\Large a+ar+ar\cdot r=26\]\[\Large a+ 6+ 6r = 26\]etc...

Answer 7

yeah nice i did it quite differently
\[a_1a_2a_3=216\\since \frac{a_2}{a_1}=\frac{a_3}{a_2}\implies a_2^2=a_1 a_3\\a_1a_2a_3=a_2a_2^2=a_2^3=216\\a_2=6\\a_1+a_2+a_3=26\implies a_1+a_3=20,a_1a_3=6^2=36\\hence\\a_1=8....a_3=2\\a_1,a_2,a_3=8,6,2\]
\[S_\infty=\frac{8}{1-3/4}\]

Answer 8

So... was there even a problem? :/

Answer 9

i dont know

Answer 10

hence
Given that the product of the first n terms of a geometric sequence is P ,find
\[\large\prod_{k=1}^na_{3k-1}\]