The first four terms of a geometric sequence are \[a, \ b, \ 4a, \ a-6+b\] for some numbers \(a\) and \(b\) with \(a\neq 0\).
Determine all possibilities for the values of \(a\) and \(b\).
Note: A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a non-zero constant, called the common ratio. For example, \(2\), \(6\), \(18\) is a geometric sequence with three terms and a common ratio of \(3\). You can explore geometric sequences further in our CEMC Courseware.
In a geometric sequence with first term \(t_1\), second term \(t_2\), third term \(t_3\), and fourth term \(t_4\), we have \(\dfrac{t_1}{t_2} = \dfrac{t_2}{t_3} = \dfrac{t_3}{t_4}\).
Here, \(t_1 = a\), \(t_2 = b\), \(t_3 = 4a\), and \(t_4 = a-6+b\).
Therefore, from \(\dfrac{t_1}{t_2} = \dfrac{t_2}{t_3}\), we have \[\begin{aligned} \frac{a}{b} &= \frac{b}{4a}\\ 4a^2 &= b^2\\ 2a &= \pm b \end{aligned}\]
There are now two cases to consider.
Case 1: When \(b = 2a\), then the sequence can be written as \(a\), \(2a\), \(4a\), \(3a-6\).
Then the ratio \(\dfrac{t_1}{t_2} = \dfrac{t_3}{t_4}\) simplifies to \[\begin{aligned} \frac{a}{2a} &= \frac{4a}{3a-6}\\ 3a^2 - 6a &= 8a^2\\ 0 & = 5a^2 + 6a\\ & = a(5a+6) \end{aligned}\]
Thus, \(a = 0\) or \(a = -\frac{6}{5}\). Since \(a \neq 0\), we have \(a = -\frac{6}{5}\).
Then \(b = 2a = 2(-\frac{6}{5}) = -\frac{12}{5}\) and \(a-6+b = -\frac{6}{5} - 6 +(-\frac{12}{5}) = -\frac{48}{5}\)
Therefore, the sequence, \(a\), \(b\), \(4a\), \(a-6+b\) is \(-\frac{6}{5}\), \(-\frac{12}{5}\), \(-\frac{24}{5}\), \(-\frac{48}{5}\).
This sequence is indeed geometric, with first term \(-\frac{6}{5}\) and common ratio \(2\).
Case 2: When \(b = -2a\), then the sequence can be written as \(a\), \(-2a\), \(4a\), \(-a-6\).
Then the ratio \(\dfrac{t_1}{t_2} = \dfrac{t_3}{t_4}\) simplifies to \[\begin{aligned} \frac{a}{-2a} &= \frac{4a}{-a-6}\\ -a^2 -6a &=-8a^2\\ 7a^2 - 6a & = 0\\ a(7a - 6) & = 0 \end{aligned}\]
Thus, \(a = 0\) or \(a = \frac{6}{7}\). Since \(a \neq 0\), we have \(a = \frac{6}{7}\).
Then \(b = -2a = -2(\frac{6}{7}) = -\frac{12}{7}\) and \(a-6+b = \frac{6}{7} - 6 +(-\frac{12}{7}) = -\frac{48}{7}\)
Therefore, the sequence, \(a, b, 4a, a-6+b\) is \(\frac{6}{7}\), \(-\frac{12}{7}\), \(\frac{24}{7}\), \(-\frac{48}{7}\).
This sequence is indeed geometric, with first term \(\frac{6}{7}\) and common ratio \(-2\).
Therefore, there are two possibilities. It could be the case that \(a=-\frac{6}{5}\) and \(b=-\frac{12}{5}\), or it could be the case that \(a=\frac{6}{7}\) and \(b=-\frac{12}{7}\).