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Consider an infinite geometric series with the first term $a$ and common ratio $r .$ If its sum is 4 and the second term is 3/4, then

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$a = 3, r = 1 / 4$

$a = 3 / 2, r = 1 / 2$

$a = 4 / 7, r = 3 / 7$

$a = 2, r = 3 / 8$

answer is D.

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Detailed Solution

We have

$\frac{a}{1 - r} = 4$ and $a r = \frac{3}{4} \Rightarrow r = \frac{3}{4 a}$

Thus, $\frac{a}{1 - 3 / 4 a} = 4 \Rightarrow a^{2} - 4 a + 3 = 0 \Rightarrow a = 1, 3$

When $a = 1, r = 3 / 4$ and when $a = 3, r = 1 / 4 .$

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