Here $g_1,g_2$ and $g_3$ are three successive terms of a Geometric Progression.
I assumed them to be $a/r, a$ and $ar$ and used $B^2 = A.C$ but proved futile...Any trick?
Asked
Active
Viewed 179 times
1
-
Geometric progression is denoted by GP – Tanishka Marrott Oct 28 '16 at 15:18
1 Answers
1
Assuming $g1=a $, $g2=ar $, $g3=ar^2$, we have $$a <0$$ and $$ar^2>4ar-3a$$ Therefore we can write $$r^2-4r+3<0$$ which has solution $$ 1 <r <3$$
Note that the possibility $a >0$ and $r <0$, assuming $g1=ar $, $g2=ar^2$, $g3=ar^3$, would lead to the same solution $ 1 <r <3\, $, which in this case contradicts the assumption $r <0$. So the solution above is the only possible one.
Anatoly
- 17,079
-
-
No, assuming $g1=ar$, $g2=ar^2$, $g3=ar^3$ with $a>0$ and $r <0$, we get $ar^3>4ar^2-3ar$ and then again $r^2-4r+3<0$. This has solution $ 1 <r <3$, which contradicts the assumption $r <0$. – Anatoly Oct 28 '16 at 17:40
-
Thus, this discussion should be part of your solution, don't you think ? – Jean Marie Oct 28 '16 at 19:36
-