Question

The sum of three consecutive positive terms of a geometric progression is S and their product is 27. Find the maximum value of S.

Answer 1

Let the terms be a/r,a,ar

Given:

Product =216

or, a/r×a×ar=216

or, a^3=216

or, a=6

And

(a/r×a)+(a×ar)+(ar×a/r)=156

or, a^2/r+a^2r+a^2=156

or, 36/r+36r+36=156

or, 36/r+36r=156−36

or, 36/r+36r=120

or, 36r^2−120r+36=0

or, 3r^2−10r+3=0

or, 3r^2−9r−r+3=0

or, 3r(r−3)−1(r−3)=0

or, (r−3)(3r−1)=0

or, r=3 or r=31

The terms of progression are

2,6,18  or  18,6,2

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Answer 2

Answer:

Let the terms be a/r,a,ar

Given:

Product =216

or, a/r×a×ar=216

or, a^3=216

or, a=6

And

(a/r×a)+(a×ar)+(ar×a/r)=156

or, a^2/r+a^2r+a^2=156

or, 36/r+36r+36=156

or, 36/r+36r=156−36

or, 36/r+36r=120

or, 36r^2−120r+36=0

or, 3r^2−10r+3=0

or, 3r^2−9r−r+3=0

or, 3r(r−3)−1(r−3)=0

or, (r−3)(3r−1)=0

or, r=3 or r=31

The terms of progression are

2,6,18 or 18,6,2

Step-by-step explanation:

@Genius