If $\frac{a}{b} = \frac{3}{4}, \frac{b}{c} = \frac{4}{5} and \frac{c}{d} = \frac{5}{6}$, then the sum of the numerator and the denominator (which are coprimes) of $\left(\frac{a}{d}\right)^{10}$ is:

1025

$\frac{a}{b} = \frac{3}{4}, \frac{b}{c} = \frac{4}{5} and \frac{c}{d} = \frac{5}{6}$

⇒ a : b : c : d = 3 : 4 : 5 : 6

⇒ (\(\frac{3}{6}\))¹⁰ = (\(\frac{1}{2}\))¹⁰

⇒ \(\frac{1}{1024}\)

⇒ Required sum of numerator and denominator = 1 + 1024 = 1025.