\(5\frac{2}{5}\)
\(4\frac{1}{5}\)
\(13\frac{1}{2}\)
\(6\frac{3}{4}\)
Correct answer is A
Sum to infinity of a G.P when /r/ < 1 = \(\frac{a}{1 - r}\)
a = \(\frac{9}{2},r = \frac{T_2}{T_1} = \frac{3}{4} + \frac{9}{2}\)
r = \(\frac{3}{4} \times \frac{2}{9} = \frac{1}{6}\)
\(S_∞ = \frac{\frac{9}{2}}{1-\frac{1}{6}} = \frac{\frac{9}{2}}{\frac{5}{6}} = \frac{27}{5}\)
\(\therefore S_∞ = 5\frac{2}{5}\)
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