\(\frac{1}{2}\)
\(\frac{3}{5}\)
\(\frac{-1}{5}\)
\(\frac{73}{12}\)
Correct answer is A
Sum to infinity
∑ = arn − 1
= \(\frac{a}{1}\) − r
a = \(\frac{1}{4}\)
r = \(\frac{1}{8}\) ÷ \(\frac{1}{4}\)
r = \(\frac{1}{s}\) × \(\frac{4}{1}\)
= \(\frac{1}{2}\)
S = \(\frac{1 \div 4}{1}\) − \(\frac{1}{2}\)
= \(\frac{1}{4}\) ÷ \(\frac{1}{2}\)
= \(\frac{1}{4}\) × \(\frac{2}{1}\)
= \(\frac{1}{2}\)