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:
Answer & explanation
Correct answer: option 1
$\frac{a}{b} = \frac{3}{4}, \frac{b}{c} = \frac{4}{5} and \frac{c}{d} = \frac{5}{6}$
⇒ 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.