Can the values of the expressions $\frac{1}{\sqrt{2a+1}},\frac{1}{\sqrt{2a-1}},3(a^2-19)\sqrt{2a-1},18\sqrt{2a+1}$ be in G.P.?

Question

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!

Answer 1

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$(-1,1)$

Answer 2

$\frac{\dfrac{1}{\sqrt{2a-1}}}{\dfrac{1}{\sqrt{2a+1}}}=\frac{18\sqrt{2a+1}}{3(a^2-19)\sqrt{2a-1}}\Rightarrow \frac{6}{a^2-19}=1$

$\frac{6}{a^2-19}=1 \Rightarrow a^2=25$ 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

Answer 3

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 \ . $