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Can the values of the expressions $\dfrac{1}{\sqrt{2a+1}},\dfrac{1}{\sqrt{2a-1}},3(a^2-19)\sqrt{2a-1},18\sqrt{2a+1}$ be in geometric progression (in the given order)?

I am confused by the fact that the expressions aren't defined for every value of $a$. Should I determine the range of $a$ or this won't be necessary in the solution? I mean we will have $$\begin{cases}2a+1>0\\2a-1>0\end{cases}\iff\begin{cases}a>-\dfrac12\\a>\dfrac12\end{cases}\iff a\in\left(\dfrac12;+\infty\right).$$ Can we use this further in the solution in some way? Thank you in advance!

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    $\begingroup$ A geometric progression has the property $\frac{a_{n+1}}{a_n} = \frac{a_n}{a_{n-1}}$. $\endgroup$ Jan 11 '21 at 16:31
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    $\begingroup$ You are expected to determine if there is any value of $a$ that makes this a geometric progression. The fact that some of the terms are not defined for some values of $a$ restricts the range of $a$ you have to consider. You are correct that if there is such an $a$ it must be greater than $\frac 12$ $\endgroup$ Jan 11 '21 at 16:33
  • $\begingroup$ @RossMillikan, thank you! Can you help me to continue? $\endgroup$
    – Katherine
    Jan 11 '21 at 16:39
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$\frac{\dfrac{1}{\sqrt{2a-1}}}{\dfrac{1}{\sqrt{2a+1}}}=\frac{18\sqrt{2a+1}}{3(a^2-19)\sqrt{2a-1}}\Rightarrow \frac{6}{a^2-19}=1$

$\frac{6}{a^2-19}=1 \Rightarrow a^2=25$ which means $a=5$ or $a=-5$

but $a=-5 \Rightarrow 2a+1<0$ so $a=5$

$\frac{3(a^2-19)\sqrt{2a-1}}{\dfrac{1}{\sqrt{2a-1}}}=3(a^2-19)(2a-1)=162 \neq 5$ which means it is a fact that values given above cannot express a geometric progression

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    $\begingroup$ You have shown that the only possible solution is $a=5$ but have not shown that it is a solution. It could be that the ratio of the third term to the second is not the same. I didn't check. $\endgroup$ Jan 11 '21 at 16:35
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Hint:

WLOG

The terms can be written as $$b,br,br^2,br^3$$

$$\implies br^3\cdot b=br\cdot br^2$$

$$18=3(a^2-19)\implies a^2=?$$

Again to keep $\sqrt{2a\pm1}$ real, $2a\ge$max$(-1,1)$

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  • $\begingroup$ I don't understand what you have tried to show me the first two lines of your answer. This is a property of G.P., isn't it? I got that $a=\pm5.$ What does this mean? Basically, we are assuming that the values are in G.P. and we are trying to find $a$, right? $\endgroup$
    – Katherine
    Jan 11 '21 at 16:49
  • $\begingroup$ Hello?................. $\endgroup$
    – Katherine
    Jan 11 '21 at 17:39
  • $\begingroup$ @nicoledobreva, In a geometric series, we can always choose $b,r$ to be the first term & the common ratio respectively, right ? Then what will be successive terms ? $\endgroup$ Jan 11 '21 at 17:46
  • $\begingroup$ I mean are we assuming that the values are in G.P., and then trying to find $a$? $\endgroup$
    – Katherine
    Jan 11 '21 at 17:51
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    $\begingroup$ As with the other answer, this is enough to demonstrate that $a=5$ is the only possibility, but not that it works. You have the ratio of the first two terms matching the ratio of the last two, but have not checked it against the ratio of the middle two. $\endgroup$ Jan 11 '21 at 20:17
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The functions of $ \ a \ $ given seem designed to mislead calculations of ratios between the functions. Selecting the pairs $$ r^2 \ \ = \ \ \frac{j(a)}{g(a)} \ \ = \ \ \frac{18 \sqrt{2a+1}}{\frac{1}{\sqrt{2a-1}}} \ \ = \ \ \frac{h(a)}{f(a)} \ \ = \ \ \frac{3·(a^2-19)· \sqrt{2a-1}}{\frac{1}{\sqrt{2a+1}}} $$ leads to $ \ 18 \sqrt{4a^2-1} \ = \ 3·(a^2-19)·\sqrt{4a^2-1} \ \Rightarrow \ a^2 = 25 \ \ . $ However, if we instead look at

$$ r^3 \ \ = \ \ \frac{j(a)}{f(a)} \ \ = \ \ 18 · (2a+1) \ \ \ \text{and} \ \ \ r \ \ = \ \ \frac{h(a)}{g(a)} \ \ = \ \ 3·(a^2-19)· (2a-1) \ \ , $$

we observe that this pair of equations is inconsistent for either $ \ a = -5 \ $ or $ \ a = 5 \ \ . $ If we solve (with the aid of WA)
$$ [ \ 3·(a^2-19)· (2a-1) \ ]^3 \ \ = \ \ 18·(2a+1) \ \ , $$ we obtain the (real) solutions $ \ a \ \approx \ -4.379 \ , \ 0.471 \ , \ 4.386 \ \ , $ two of which are excluded as we shall see. So this would indicate that a geometric progression is not possible among the functions.

In fact the properties of the functions themselves tell us that this will fail to occur. The common domain of all four functions is $ \ a > \frac12 \ \ $ and $ \ h(a) \ = \ 3·(a^2-19)· \sqrt{2a-1} \ $ is not positive on $ \ [ \ \frac12 \ , \ \sqrt{19} \ ] \ \ . $ Moreover, $ \ h(a) \ = \ j(a) \ $ for $ \ a_0 \ \approx \ 5.06 \ \ , $ so the only plausible "window" for which these functions might be in geometric progression is $ \ \sqrt{19} < a < a_0 \ \ . $ We find that in this interval that $ \ j(a) \ $ has values around $ \ 55 \ $ to $ \ 60 \ \ , $ while $ \ f(a) \ $ and $ \ g(a) \ $ are very close together near $ \ 0.31 \ $ and $ \ 0.34 \ \ . $ So there is no value that $ \ h(a) \ $ can take on in this interval that will link the four functions in a geometric progression; the order of the functions changes, but the situation does not for $ \ a > a_0 \ . $

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