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

Given, $5, 5r, 5{r}^{2}$5,5r,5r2 are the length of sides of triangle, then we know that the sum of two sides of a triangle is more than the third side i.e.

$$5+5r>5{r}^{2}...(1)$$5+5r>5r2...(1)

$$5+5{r}^{2}>5r...(2)$$5+5r2>5r...(2)

$$5r+5{r}^{2}>5...(3)$$5r+5r2>5...(3)

From $(1),$(1),

$${r}^{2}-r-1<0$$r2−r−1<0

$$[r-(\frac{1+\sqrt{5}}{2})] [r-(\frac{1-\sqrt{5}}{2}) ]<0$$[r−(21+5​​)][r−(21−5​​)]<0

$$r\in (\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2})...(4)$$r∈(21−5​​,21+5​​)...(4)

from $(2),$(2),

$${r}^{2}-r+1>0$$r2−r+1>0

The discriminant of the quadratic is $D={(-1)}^{2}-4\times 1\times 1=-3$D=(−1)2−4×1×1=−3 and we know that if the discriminant of a quadratic is negative and its leading coefficient is positive then the quadratic is positive for all real numbers.

$$\Rightarrow r\in R...(5)$$⇒r∈R...(5)

from $(3),$(3),

$${r}^{2}+r-1>0$$r2+r−1>0

$$(r+\frac{1+\sqrt{5}}{2})(r+\frac{1-\sqrt{5} }{2})>0$$(r+21+5​​)(r+21−5​​)>0

So, $r\in (-\infty , -\frac{1+\sqrt{5}}{2})\cup (-\frac{1-\sqrt{5}}{2}, \infty )...(6)$r∈(−∞,−21+5​​)∪(−21−5​​,∞)...(6)

Now, taking intersection of $(4),(5),(6),$(4),(5),(6), we get $r\in (\frac{-1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}).$r∈(2−1+5​​,21+5​​).

Out of the given options only $\frac{7}{4}$47​ is not in the interval obtained.