Question

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: Relate \(a_2\) and \(a_4\)

We know that \(a_2 = 20\) and \(a_4=\frac{45}{4}\). Since \(a_4=a_1r^{3}\) and \(a_2 = a_1r\), we can express \(a_4\) as \(a_4=a_2\times r^{2}\) (because \(a_1r^{3}=(a_1r)\times r^{2}=a_2r^{2}\)).

Step3: Solve for \(r\)

Substitute \(a_2 = 20\) and \(a_4=\frac{45}{4}\) into \(a_4=a_2r^{2}\), we get \(\frac{45}{4}=20\times r^{2}\). Then, divide both sides by 20: \(r^{2}=\frac{45}{4}\div20=\frac{45}{4}\times\frac{1}{20}=\frac{45}{80}=\frac{9}{16}\). Take the square root of both sides: \(r=\pm\sqrt{\frac{9}{16}}=\pm\frac{3}{4}\). But in a geometric sequence, if we consider the terms \(a_2 = 20\) and \(a_4=\frac{45}{4}\) (which is positive), the common ratio is likely positive (since if \(r\) is negative, \(a_3\) would be negative and \(a_4\) would be positive, but \(a_2\) is positive, it's possible, but let's check the values. If \(r =-\frac{3}{4}\), then \(a_3=a_2r=20\times(-\frac{3}{4})=- 15\), \(a_4=a_3r=-15\times(-\frac{3}{4})=\frac{45}{4}\), which also works. But usually, we consider the positive ratio first, and both are valid mathematically. But let's see the calculation again. Wait, when we did \(a_4 = a_2r^{2}\), so \(r^{2}=\frac{45}{4\times20}=\frac{9}{16}\), 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 matches. If \(r =-\frac{3}{4}\), \(a_3=20\times(-\frac{3}{4})=-15\), \(a_4=-15\times(-\frac{3}{4})=\frac{45}{4}\), also matches. But maybe the problem expects the positive ratio, so \(r = \frac{3}{4}\) or \(r=-\frac{3}{4}\). But let's see the problem statement, it just says "common ratio", so both are possible, but let's check the arithmetic. Wait, maybe I made a mistake in step 2? Wait, \(a_4=a_1r^3\), \(a_2=a_1r\), so \(a_4=a_2r^{2}\), that's correct. So \(r^{2}=\frac{45/4}{20}=\frac{45}{80}=\frac{9}{16}\), so \(r=\pm\frac{3}{4}\). But let's see the values, if we take \(r=\frac{3}{4}\), the sequence is positive - decreasing, which fits \(a_2 = 20\) and \(a_4=\frac{45}{4}\). So the common ratio is \(\frac{3}{4}\) (or \(-\frac{3}{4}\), but likely \(\frac{3}{4}\) as the problem gives positive terms for \(a_2\) and \(a_4\) (even though \(a_4\) is a fraction, it's positive)).

Answer:

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