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If sum of an infinite geometric series is $\frac{4}{5}$ and its 1 st term is $\frac{3}{4}$, then its common ratio is
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The correct answer is:
$\frac{7}{16}$
Hints: $\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{4}{3} \quad$ Then $\frac{\frac{3}{4}}{1-\mathrm{r}}=\frac{4}{3} \Rightarrow \mathrm{r}=1-\frac{9}{16}=\frac{7}{16}$
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