Question

question 3 of 10
the second term in a geometric sequence is 20. the fourth term in the same sequence is 45/4, or 11.25. what is the common ratio in this sequence?

Explanation:

Step1: Recall geometric sequence formula

In a geometric sequence, the \(n\)-th term is given by \(a_n = a_1r^{n - 1}\), where \(a_1\) is the first term and \(r\) is the common ratio. So, the second term \(a_2=a_1r^{2 - 1}=a_1r\), and the fourth term \(a_4=a_1r^{4 - 1}=a_1r^{3}\).

Step2: Divide the formula for \(a_4\) by \(a_2\)

We know \(a_2 = 20\) and \(a_4=\frac{45}{4}\). Divide \(a_4\) by \(a_2\): \(\frac{a_4}{a_2}=\frac{a_1r^{3}}{a_1r}\). The \(a_1\) terms cancel out, and using the rule of exponents \(\frac{r^{m}}{r^{n}}=r^{m - n}\), we get \(\frac{a_4}{a_2}=r^{3 - 1}=r^{2}\).
Substitute the values of \(a_4\) and \(a_2\): \(\frac{\frac{45}{4}}{20}=r^{2}\). Calculate \(\frac{\frac{45}{4}}{20}=\frac{45}{4}\times\frac{1}{20}=\frac{45}{80}=\frac{9}{16}\). So, \(r^{2}=\frac{9}{16}\).

Step3: Solve for \(r\)

Take the square root of both sides: \(r=\pm\sqrt{\frac{9}{16}}\). \(\sqrt{\frac{9}{16}}=\frac{3}{4}\), so \(r = \pm\frac{3}{4}\). But in a geometric sequence, if we consider the terms \(a_2 = 20\) and \(a_4=\frac{45}{4}\) (which is positive), if \(r\) were negative, \(a_3=a_2r\) would be negative and \(a_4=a_3r\) would be positive, but let's check the magnitude. Let's test \(r=\frac{3}{4}\): \(a_3=a_2r = 20\times\frac{3}{4}=15\), \(a_4=a_3r=15\times\frac{3}{4}=\frac{45}{4}\), which matches. If \(r = -\frac{3}{4}\), \(a_3=20\times(-\frac{3}{4})=- 15\), \(a_4=-15\times(-\frac{3}{4})=\frac{45}{4}\), which also gives the correct \(a_4\). But usually, we consider the positive ratio for the main sequence (unless specified otherwise, and the terms here can have a positive ratio as the terms go from 20 to a smaller positive number, so the common ratio is likely positive. So we take \(r=\frac{3}{4}\) (or we can also consider the negative, but in the context of a sequence where the second term is positive and the fourth term is positive, both positive and negative ratios are mathematically valid, but let's see the calculation again. Wait, when we did \(\frac{a_4}{a_2}=r^{2}\), \(\frac{45/4}{20}=\frac{45}{80}=\frac{9}{16}\), so \(r^{2}=\frac{9}{16}\), so \(r=\pm\frac{3}{4}\). But let's check with \(r = \frac{3}{4}\): \(a_1=\frac{a_2}{r}=\frac{20}{\frac{3}{4}}=\frac{80}{3}\), \(a_3=\frac{80}{3}\times(\frac{3}{4})^2=\frac{80}{3}\times\frac{9}{16}=15\), \(a_4 = 15\times\frac{3}{4}=\frac{45}{4}\), which is correct. If \(r=-\frac{3}{4}\), \(a_1=\frac{20}{-\frac{3}{4}}=-\frac{80}{3}\), \(a_3=-\frac{80}{3}\times\frac{9}{16}=-15\), \(a_4=-15\times(-\frac{3}{4})=\frac{45}{4}\), also correct. But since the problem doesn't specify the sign, but the terms are positive, and usually, we take the positive ratio for such problems (as a negative ratio would alternate signs, but here \(a_2\) and \(a_4\) are both positive, so \(r\) can be positive or negative, but let's see the calculation again. Wait, the key is that when we divide \(a_4\) by \(a_2\), we get \(r^{2}=\frac{9}{16}\), so \(r=\pm\frac{3}{4}\). But let's check with \(r = \frac{3}{4}\): \(a_2 = 20\), \(a_3=20\times\frac{3}{4}=15\), \(a_4=15\times\frac{3}{4}=\frac{45}{4}\), which is correct. So the common ratio is \(\frac{3}{4}\) (or \(-\frac{3}{4}\), but likely \(\frac{3}{4}\) as the terms are decreasing in magnitude and positive).

Answer:

\(\frac{3}{4}\) (or \(-\frac{3}{4}\), but the positive value is more probable in this context)