The first term in a geometric sequence is \(a\), the second term is \(b\), and the third term is \(c\). The three terms have a sum of \(158\) and a product of \(74\,088\).
Determine all possible ordered triples \((a,b,c)\).
Note: The general term of a geometric sequence can be written as \(t_n=ar^{n-1}\), where \(a\) is first term of the sequence, \(r\) is the common ratio between terms, and \(t_n\) is the \(n^{th}\) term.
Let \(r\) be the common ratio of the geometric sequence. Since \(a\) is the first term of the sequence, then \(b=ar\) and \(c=ar^2\).
We are given that \(abc=74\,088\). Thus, \(a(ar)(ar^2)=a^3r^3 = (ar)^3=74\,088\). Therefore, \(ar=42\). Since \(b=ar\), we have \(b=42\).
Now, \(a+b+c=158\) becomes \(a+42+c=158\), or \(a+c=116\).
Since \(b=ar\), then \(42=ar\), or \(r = \dfrac{42}{a}\) (since the product of \(a\), \(b\), and \(c\) is not zero, we know \(a\neq 0\)).
Therefore, \(c=ar^2 = a\left(\dfrac{42}{a}\right)^2= a\left(\dfrac{1764}{a^2}\right)=\dfrac{1764}{a}\).
Substituting \(c = \dfrac{1764}{a}\) into \(a+c = 116\), we have \[\begin{aligned} a+ \frac{1764}{a} &= 116\\ a^2 +1764 &= 116a\\ a^2 -116a+1764&=0\\ (a-18)(a-98)&=0 \end{aligned}\]
Therefore, \(a=18\) or \(a=98\).
When \(a = 18\), then \(r = \frac{42}{18}=\frac{7}{3}\), and one
ordered triple is \((18,\,42,\,98)\).
Indeed, we can check that \(18 + 42 + 98 =
158\) and \((18)(42)(98) =
74\,088\).
When \(a=98\), then \(r = \frac{42}{98}=\frac{3}{7}\), and one
ordered triple is \((98,\,42,\,18)\).
Indeed, we can check that \(98 + 42 + 18 =
158\) and \((98)(42)(18) =
74\,088\).
In conclusion there are two ordered triples that satisfy the conditions of the problem. They are \((18,\,42,\,98)\) and \((98,\,42,\,18)\).