If 5, 5$$r$$, 5$$r^2$$ are the lengths of the sides of a triangle, then $$r$$ can not be equal to:

We are told that the three sides of a triangle have lengths $$5,\;5r,\;5r^{2}$$, where $$r\gt 0$$. For any three positive numbers to be the sides of a triangle, they must satisfy the triangle‐inequality theorem, which states:

Sum of any two sides > the remaining side.

Because the factor $$5$$ is common to all three expressions, it is convenient to divide each inequality by $$5$$. This leaves the simpler numbers $$1,\;r,\;r^{2}$$. Thus, the triangle inequalities become:

$$\begin{aligned} 1+r \gt r^{2},\\[4pt] r+r^{2} \gt 1,\\[4pt] 1+r^{2} \gt r. \end{aligned}$$

The third inequality $$1+r^{2}\gt r$$ is automatically true for every positive $$r,$$ so only the first two need detailed analysis. We examine them by separating the two natural cases $$r\ge 1$$ and $$0\lt r\le 1.$$

Case 1: $$r\ge 1.$$} In this range $$r^{2}\ge r\ge 1,$$ so $$r^{2}$$ is the largest of the three numbers $$1,\;r,\;r^{2}.$$ The pertinent inequality is therefore

$$1+r\gt r^{2}.$$

Rearranging gives

$$r^{2}-r-1\lt 0.$$

We solve the quadratic $$r^{2}-r-1=0$$ by the quadratic formula:

$$r=\frac{1\pm\sqrt{1+4}}{2}=\frac{1\pm\sqrt5}{2}.$$

Thus the two real roots are $$\dfrac{1-\sqrt5}{2}\;(\text{negative})$$ and $$\dfrac{1+\sqrt5}{2}\approx1.618.$$ A quadratic with positive leading coefficient is negative between its roots, so

$$-\,\frac{\sqrt5-1}{2}\lt r\lt \frac{1+\sqrt5}{2}.$$

Because we are in the subcase $$r\ge1,$$ the combined condition is

$$1\le r\lt \frac{1+\sqrt5}{2}\quad\Longrightarrow\quad1\le r\lt 1.618\ldots$$

Case 2: $$0\lt r\le1.$$} Here the largest of the three numbers $$1,\;r,\;r^{2}$$ is $$1.$$ Hence we must satisfy

$$r+r^{2}\gt 1.$$

Rewriting yields

$$r^{2}+r-1\gt 0.$$

We solve the quadratic $$r^{2}+r-1=0$$ in the same way:

$$r=\frac{-1\pm\sqrt{1+4}}{2}=\frac{-1\pm\sqrt5}{2}.$$

The positive root is $$\dfrac{\sqrt5-1}{2}\approx0.618.$$ Because the quadratic opens upward, it is positive outside the interval between its roots, giving

$$r\gt \frac{\sqrt5-1}{2}\quad\text{or}\quad r\lt -\frac{1+\sqrt5}{2}.$$

Only positive $$r$$ are relevant, so for this case

$$\frac{\sqrt5-1}{2}\lt r\le1\quad\Longrightarrow\quad0.618\ldots\lt r\le1.$$

Combining both cases we see that a valid $$r$$ must lie in the single continuous interval

$$0.618\ldots\lt r\lt 1.618\ldots$$

Now we test the four given choices:

$$\begin{aligned} \frac34 &=0.75 &\checkmark\\[4pt] \frac32 &=1.50 &\checkmark\\[4pt] \frac54 &=1.25 &\checkmark\\[4pt] \frac74 &=1.75 &\times \end{aligned}$$

The only option that lies outside the required range is $$\dfrac74.$$ Therefore $$r$$ cannot be equal to $$\dfrac74.$$ Hence, the correct answer is Option D.